JEE Main Physics — Electrostatics & Current Electricity Complete Notes

Part I: Electrostatics

Introduction: The Foundation of Modern Technology

Electrostatics is not merely a chapter in your JEE syllabus; it is the cornerstone of classical electromagnetism. Every electronic device you use, from the capacitive touchscreen of your smartphone to the high-voltage transmission lines powering your city, operates on the principles defined here. For a JEE Main aspirant, this unit is high-yield, typically accounting for 3–4 direct questions ($12–16$ marks). However, its importance extends further, as the concepts of potential, field vectors, and flux are mirrored in Gravitation and Magnetism.

In this resource, we transition from the macroscopic observation of "rubbing amber" to the microscopic reality of electron transfer. We will dissect the subtle nuances between Conductors and Insulators, the geometry of field lines, and the rigorous mathematical frameworks of Coulomb’s Law and Gauss’s Theorem. Success in JEE Physics requires a "dual-lens" approach: first, a qualitative intuition to visualize the field, and second, a quantitative precision to solve complex vector summations. We begin our journey with the most fundamental entity in the universe: the Electric Charge.

1. Properties of Electric Charge

Charge is an intrinsic property of matter, responsible for electromagnetic interactions.

1.1 Additivity of Charge

Total charge of a system is the algebraic sum of individual charges. $$Q_{total} = q_1 + q_2 + q_3 + \dots + q_n$$ Note: Signs (+/-) must be included.

1.2 Conservation of Charge

In an isolated system, the total charge remains constant. Charge can neither be created nor destroyed; it can only be transferred from one body to another (e.g., via friction, conduction, or induction).

1.3 Quantization of Charge

Charge exists only in discrete packets (quanta). The smallest unit of free charge is the electronic charge $e \approx 1.6 \times 10^{-19} \text{ C}$. $$Q = \pm ne$$ Where $n$ is an integer ($n = 1, 2, 3, \dots$).

2. Coulomb’s Law

Describes the electrostatic force between two stationary point charges.

2.1 Scalar Form

The magnitude of the force ($F$) between two point charges $q_1$ and $q_2$ separated by distance $r$: $$F = k \frac{|q_1 q_2|}{r^2}$$ Where $k$ is the electrostatic constant: $$k = \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \text{ N m}^2/\text{C}^2$$ $\epsilon_0$ (Permittivity of free space) $\approx 8.854 \times 10^{-12} \text{ C}^2/\text{N m}^2$.

2.2 Vector Form

Force on $q_2$ due to $q_1$: $$\vec{F}{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^3} \vec{r}{21}$$ Where $\vec{r}_{21}$ is the position vector from $q_1$ to $q_2$.

2.3 Principle of Superposition

The net force on a charge is the vector sum of forces exerted by all other individual charges. $$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \dots + \vec{F}_n$$

Worked Example 1: Equilibrium of Charges Question: Two point charges $+q$ and $+4q$ are separated by a distance $L$. Where should a third charge $Q$ be placed so that it is in equilibrium? Solution: For $Q$ to be in equilibrium, the forces from $+q$ and $+4q$ must be equal and opposite. Let $x$ be the distance from $+q$. $$k\frac{qQ}{x^2} = k\frac{4qQ}{(L-x)^2}$$ $$\frac{1}{x^2} = \frac{4}{(L-x)^2} \implies \frac{1}{x} = \frac{2}{L-x}$$ $$L-x = 2x \implies 3x = L \implies x = L/3$$ Answer: $L/3$ from the $+q$ charge.

3. Electric Field ($E$)

The electric field at a point is the force experienced by a unit positive test charge ($q_0$) placed at that point. $$\vec{E} = \lim_{q_0 \to 0} \frac{\vec{F}}{q_0}$$ Units: N/C or V/m.

3.1 Field due to various geometries

A. Point Charge

$$E = \frac{kq}{r^2}$$

B. Infinite Line Charge (Linear charge density $\lambda$)

Using Gauss Law (to be detailed later): $$E_{line} = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2k\lambda}{r}$$ Direction: Radially outward if $\lambda > 0$.

C. Infinite Non-conducting Sheet (Surface charge density $\sigma$)

$$E_{sheet} = \frac{\sigma}{2\epsilon_0}$$ Note: The field is independent of the distance from the sheet.

D. Uniformly Charged Ring

Field at a point on the axis at distance $x$ from center (radius $R$): $$E_{ring} = \frac{kqx}{(R^2 + x^2)^{3/2}}$$ Max field occurs at $x = R/\sqrt{2}$.

E. Uniformly Charged Disc

Surface density $\sigma$, radius $R$, distance $x$ on axis: $$E_{disc} = \frac{\sigma}{2\epsilon_0} \left[ 1 - \frac{x}{\sqrt{R^2 + x^2}} \right]$$

F. Uniformly Charged Spherical Shell (Radius $R$)

1. Outside ($r > R$): $E = \frac{kq}{r^2}$ (Acts like a point charge at center).
2. On Surface ($r = R$): $E = \frac{kq}{R^2}$.
3. Inside ($r < R$): $E = 0$ (Zero field inside a conductor/shell).

4. Electric Dipole

A system of two equal and opposite charges $\pm q$ separated by a small distance $2l$.

4.1 Dipole Moment ($\vec{p}$)

$$\vec{p} = q \times (2\vec{l})$$ Direction: From negative charge to positive charge.

4.2 Electric Field of a Dipole

For a point at distance $r$ from the center ($r \gg l$):

Worked Example 2: Ratio of Fields Question: Find the ratio of the electric field at an axial point to the equatorial point at the same distance $r$ from a short dipole. Solution: $E_{axial} = \frac{2kp}{r^3}$ and $E_{eq} = \frac{kp}{r^3}$. Ratio = $\frac{2kp/r^3}{kp/r^3} = 2:1$.

5. Dipole in a Uniform External Field

5.1 Torque ($\vec{\tau}$)

A dipole in a uniform field $\vec{E}$ experiences a torque that tends to align it with the field. $$\vec{\tau} = \vec{p} \times \vec{E}$$ $$\tau = pE \sin \theta$$

• Max Torque: $\theta = 90^\circ \implies \tau = pE$
• Min Torque: $\theta = 0^\circ$ or $180^\circ \implies \tau = 0$

5.2 Potential Energy ($U$)

The work done in rotating a dipole in an external field: $$U = -\vec{p} \cdot \vec{E} = -pE \cos \theta$$

Worked Example 3: Work Done in Rotation Question: How much work is required to rotate a dipole $\vec{p}$ from stable equilibrium to unstable equilibrium in a field $E$? Solution: $W = U_{final} - U_{initial}$ $W = (-pE \cos 180^\circ) - (-pE \cos 0^\circ)$ $W = (pE) - (-pE) = 2pE$.

6. Electric Field Lines

Imaginary lines representing the direction of the electric field.

Key Properties:

1. They originate from positive charges and terminate on negative charges.
2. They do not form closed loops (Conservative nature of field).
3. The tangent at any point gives the direction of $\vec{E}$.
4. Two lines never intersect (direction at a point must be unique).
5. Number of lines per unit area represents the magnitude of the field.
6. They are always perpendicular to the surface of a conductor.

7. Numerical Practice and Problem-Solving

Worked Example 4: Vector Addition of Fields

Question: Three charges $+q$, $+q$, and $-q$ are placed at the vertices of an equilateral triangle of side $a$. Find the net electric field at the centroid. Solution: Let the centroid be $O$. Distance of centroid from vertex $r = a/\sqrt{3}$.

1. $\vec{E}_A$ (due to $+q$ at A) is away from A.
2. $\vec{E}_B$ (due to $+q$ at B) is away from B.
3. $\vec{E}C$ (due to $-q$ at C) is toward C. The magnitude of each individual field is $E = \frac{kq}{(a/\sqrt{3})^2} = \frac{3kq}{a^2}$. The vector sum of $\vec{E}A$ and $\vec{E}B$ points toward C with magnitude $2E \cos 60^\circ = E$. Total Field $E{net} = E{AB} + E_C = E + E = 2E$. $E{net} = \frac{6kq}{a^2}$ directed toward the $-q$ charge.

Worked Example 5: Continuous Charge Distribution

Question: A semicircular wire of radius $R$ carries a total charge $Q$ distributed uniformly. Find the electric field at the center of curvature. Solution: Linear charge density $\lambda = \frac{Q}{\pi R}$. Consider a small element $dl = R d\theta$. Charge $dq = \lambda R d\theta$. The horizontal components of $dE$ cancel due to symmetry. Vertical components add up. $dE_y = \frac{k dq}{R^2} \sin \theta = \frac{k (\lambda R d\theta)}{R^2} \sin \theta$. Integrating from $0$ to $\pi$: $E = \int_0^\pi \frac{k\lambda}{R} \sin \theta d\theta = \frac{k\lambda}{R} [-\cos \theta]_0^\pi$ $E = \frac{k\lambda}{R} [1 - (-1)] = \frac{2k\lambda}{R}$. Substituting $\lambda = Q/(\pi R)$: $E = \frac{2kQ}{\pi R^2}$.

Worked Example 6: Force between Dipoles

Question: Two short dipoles with dipole moments $p_1$ and $p_2$ are placed along the same line at a large distance $r$. What is the force between them? Solution: Force $F = |p_2 \frac{dE_1}{dr}|$, where $E_1$ is the field of $p_1$ at the location of $p_2$. For axial position, $E_1 = \frac{2kp_1}{r^3}$. $\frac{dE_1}{dr} = \frac{d}{dr}(\frac{2kp_1}{r^3}) = - \frac{6kp_1}{r^4}$. $F = p_2 \cdot \frac{6kp_1}{r^4} = \frac{6kp_1 p_2}{r^4}$. Note: This force follows an inverse-fourth power law ($1/r^4$), much shorter-range than point-charge interactions.

8. Summary Table of Formulae (The "Shortcut" Sheet)

9. Mnemonic for JEE Preparation

"Q-C-A" (Properties):

• Quantized ($ne$)
• Conserved (Closed system)
• Additive (Algebraic sum)

"S-U-T" (Dipole Energy):

• Stable: $0^\circ$ (Low energy)
• Unstable: $180^\circ$ (High energy)
• Torque: $pE \sin \theta$

10. Common Pitfalls to Avoid

1. Vector Direction: Never calculate force/field without sketching the vector directions first. The signs of charges are for direction, not just magnitude.
2. Dimension Check: Always check if your formula yields $N/C$. For example, the ring field has $x$ on top and $r^3$ on bottom, resulting in $1/r^2$ units—this is correct.
3. Shell vs Sphere: Remember that for a conducting sphere, the field inside is always zero. This is not true for a non-conducting solid sphere (to be covered in Gauss Law applications).
4. Dipole Distance: The $1/r^3$ approximation for dipoles is only valid when $r \gg l$. If $r$ is comparable to $l$, use the full formula: $E_{axial} = \frac{2kpr}{(r^2-l^2)^2}$.

Gauss's Law & Applications

Electrostatics in JEE Main is heavily centered around the concept of Flux and Gauss's Law. While Coulomb’s Law is the fundamental law of force, Gauss’s Law provides a powerful, high-symmetry alternative to calculating Electric Fields ($\vec{E}$) without undergoing rigorous integration.

1. Electric Flux ($\Phi$)

Electric flux is a measure of the "number of field lines" passing through a given surface area. Quantitatively, it is the dot product of the Electric Field vector and the Area vector.

Definition: For a flat surface in a uniform electric field: $$\Phi = \vec{E} \cdot \vec{A} = EA \cos \theta$$ where $\theta$ is the angle between the Electric Field $\vec{E}$ and the Normal to the surface.

General Case (Variable Field/Curved Surface): $$\Phi = \oint \vec{E} \cdot d\vec{A}$$

2. Gauss's Law: Statement and Significance

Statement: The net electric flux through any closed surface (called a Gaussian Surface) is equal to $1/\epsilon_0$ times the net charge enclosed within that surface.

$$\Phi_{net} = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}$$

Key Properties for JEE:

1. Independence of Shape: Flux depends only on the charge enclosed, not the shape or size of the Gaussian surface.
2. Location of Charge: It does not matter where the charge is inside the surface.
3. External Charges: Charges outside the surface contribute zero to the net flux, as every field line entering must also exit.
4. Superposition: The $\vec{E}$ in the formula is the net electric field due to all charges (inside and outside), but the flux total only equals the internal charge.

3. Derivation of Fields using Symmetry

To solve $E$ using Gauss's Law, we choose a Gaussian surface where $E$ is either constant or zero on the surface, and $\vec{E}$ is either parallel or perpendicular to $d\vec{A}$.

A. Infinite Line Charge ($\lambda$)

Consider a long wire with linear charge density $\lambda$ ($C/m$).

• Gaussian Surface: A cylinder of radius $r$ and length $l$ coaxial with the wire.
• Flux calculation: Flux through the top and bottom caps is zero (since $\vec{E} \perp d\vec{A}$). Flux exists only through the curved surface. $$\int \vec{E} \cdot d\vec{A} = E(2\pi r l)$$
• Applying Gauss Law: $$E(2 \pi r l) = \frac{\lambda l}{\epsilon_0} \implies E = \frac{\lambda}{2\pi \epsilon_0 r}$$

B. Infinite Non-Conducting Sheet ($\sigma$)

Consider a thin infinite plane sheet with surface charge density $\sigma$ ($C/m^2$).

• Gaussian Surface: A pillbox (cylinder) penetrating the sheet.
• Flux calculation: Flux through the curved surface is zero. Flux through the two end caps (area $A$ each) is $2EA$.
• Applying Gauss Law: $$2EA = \frac{\sigma A}{\epsilon_0} \implies E = \frac{\sigma}{2\epsilon_0}$$
• Note: The field is independent of the distance $r$.

C. Field Between Two Large Parallel Sheets

If we have two sheets with densities $+\sigma$ and $-\sigma$:

1. Region Outside: $E_{net} = E_+ - E_- = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$.
2. Region Inside: $E_{net} = E_+ + E_- = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$.

D. Thin Spherical Shell (Radius $R$, Total Charge $Q$)

• Outside ($r > R$): Choose a spherical Gaussian surface of radius $r$. $$E(4\pi r^2) = \frac{Q}{\epsilon_0} \implies E_{out} = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$$
• Inside ($r < R$): Since $q_{enclosed} = 0$, the field $E_{in} = 0$.

E. Uniformly Charged Solid Non-Conducting Sphere (Radius $R$, Charge $Q$)

Here, charge is distributed throughout the volume with density $\rho = \frac{Q}{\frac{4}{3}\pi R^3}$.

• Outside ($r \geq R$): Behaves like a point charge. $E = \frac{kQ}{r^2}$.
• Inside ($r < R$): $$q_{enc} = \rho \times \frac{4}{3}\pi r^3 = Q \frac{r^3}{R^3}$$ Applying Gauss Law: $E(4\pi r^2) = \frac{Q r^3}{\epsilon_0 R^3} \implies E_{in} = \frac{1}{4\pi\epsilon_0} \frac{Qr}{R^3} = \frac{\rho r}{3\epsilon_0}$

F. Conducting Sphere (Solid or Hollow)

In a conductor, all excess charges reside on the outer surface. Thus, for a solid conducting sphere, the field distribution is identical to a thin spherical shell:

• $E_{in} = 0$
• $E_{out} = \frac{kQ}{r^2}$

4. Summary Table of Electric Fields

5. 10 Solved Problems (JEE Main Pattern)

Q1. A charge $q$ is placed at the center of the open face of a hemispherical bowl. Find the flux through the bowl.

• Solution: To enclose the charge, imagine another identical hemisphere on top. Total flux through the full sphere is $q/\epsilon_0$. By symmetry, flux through one hemisphere is $\Phi = \frac{q}{2\epsilon_0}$.

Q2. A cube of side $a$ has a point charge $q$ at one of its corners. Find the flux through one of the faces touching that corner.

• Solution: The field lines from the charge are parallel to the three faces meeting at that corner ($E \perp d\vec{A}$ is false, rather $\vec{E} \parallel$ surface). Thus, $\vec{E} \cdot d\vec{A} = 0$. Flux through those three faces is zero. Total flux through the cube (by symmetry of 8 cubes) is $q/8\epsilon_0$. This flux is shared by the 3 faces opposite to the corner. So, flux through one opposite face is $(q/8\epsilon_0)/3 = q/24\epsilon_0$.

Q3. An infinitely long wire has a linear charge density $0.5 , \mu C/m$. Find the electric field at a distance of $2 , m$.

• Solution: Use $E = \frac{\lambda}{2\pi\epsilon_0 r}$. $E = \frac{0.5 \times 10^{-6}}{2 \times \pi \times 8.85 \times 10^{-12} \times 2}$ $E = \frac{2 \times 9 \times 10^9 \times 0.5 \times 10^{-6}}{2} = 4.5 \times 10^3 , V/m$.

Q4. A solid sphere of radius $R$ has a volume charge density $\rho = \rho_0 r$. Find the electric field at $r < R$.

• Solution: $q_{enc} = \int_0^r \rho (4\pi x^2) dx = \int_0^r \rho_0 x (4\pi x^2) dx = 4\pi \rho_0 \frac{r^4}{4} = \pi \rho_0 r^4$. Apply Gauss Law: $E(4\pi r^2) = \frac{\pi \rho_0 r^4}{\epsilon_0} \implies E = \frac{\rho_0 r^2}{4\epsilon_0}$.

Q5. Two infinite parallel sheets have charge densities $\sigma$ and $-2\sigma$. Find the field in the region between them.

• Solution: $E_{net} = E_{\sigma} + E_{-2\sigma} = \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$. They add up because the vectors both point from the positive sheet toward the negative sheet.

Q6. A point charge $Q$ is placed at a distance $a/2$ directly above the center of a square of side $a$. Find the flux through the square.

• Solution: Imagine the square is the base of a cube of side $a$. The charge is then at the center of the cube. Total flux $= Q/\epsilon_0$. Since the cube has 6 faces, flux through one face (the square) is $\Phi = Q/6\epsilon_0$.

Q7. If the electric field is given by $\vec{E} = 3i + 4j$, find the flux through a surface of area $100$ units in the $XY$ plane.

• Solution: Area vector for $XY$ plane is in $Z$ direction ($k$). So $\vec{A} = 100 \hat{k}$. $\Phi = \vec{E} \cdot \vec{A} = (3\hat{i} + 4\hat{j}) \cdot (100\hat{k}) = 0$.

Q8. A conducting sphere of radius $10 cm$ has an unknown charge. If the electric field $20 cm$ from the center is $1.5 \times 10^3 N/C$ radially inward, what is the net charge?

• Solution: $E = \frac{kQ}{r^2} \implies 1.5 \times 10^3 = \frac{9 \times 10^9 \times Q}{(0.2)^2}$. $Q = \frac{1.5 \times 10^3 \times 0.04}{9 \times 10^9} = 6.67 \times 10^{-9} C$. Since field is inward, $Q = -6.67 , nC$.

Q9. Find the ratio of electric field at $r = R/2$ and $r = 2R$ for a solid non-conducting uniformly charged sphere.

• Solution: $E_{in}(R/2) = \frac{kQ(R/2)}{R^3} = \frac{kQ}{2R^2}$. $E_{out}(2R) = \frac{kQ}{(2R)^2} = \frac{kQ}{4R^2}$. Ratio $= \frac{1/2}{1/4} = 2:1$.

Q10. A hollow cylindrical box of length $1m$ and area of cross-section $25 cm^2$ is placed in a three-dimensional coordinate system. The electric field is $\vec{E} = 50x \hat{i}$. Find the net charge inside.

• Solution: Flux only through caps. Left cap at $x=0$, $E=0 \implies \Phi_L = 0$. Right cap at $x=1$, $E=50(1) = 50$. $\Phi_R = EA = 50 \times 25 \times 10^{-4} = 0.125$. $\Phi_{net} = 0.125 = q/\epsilon_0 \implies q = 0.125 \times 8.85 \times 10^{-12} \approx 1.1 \times 10^{-12} C$.

6. JEE Main Pitfalls & Tips

• Pitfall 1: Distinguishing Conducting vs. Non-conducting: Always check if the sphere is a conductor. In a conductor, $E_{in}$ is always $0$. In a non-conductor, $E \propto r$.

• Pitfall 2: Flux through a Disc vs. Square: If a charge is near a disc, don't assume $\Phi = Q/6\epsilon_0$. That only works for symmetric segments of a cube. For a disc, you must use solid angle: $\Phi = \frac{q}{2\epsilon_0}(1 - \cos \alpha)$, where $\alpha$ is the semi-vertical angle.

• Pitfall 3: The "Net Field" Concept: In Gauss's Law $\oint \vec{E} \cdot d\vec{A}$, the $\vec{E}$ is the field due to all charges, but the $q$ is only the enclosed charge. JEE often tests this conceptual distinction in multiple-choice questions.

• Mnemonic for $E$ fields:

  • Line: $\lambda / 2 \pi \epsilon_0 r$ (Length involves $2\pi r$ circumference).
  • Sheet: $\sigma / 2 \epsilon_0$ (Sheet has 2 sides, hence the '2' in denominator).
  • Sphere (In): $r$ (Inside is Rinear... well, linear).
  • Sphere (Out): $1/r^2$ (Outside, it acts like a Point).

• Tip for Cavity Problems: If a cavity is created inside a solid sphere, the field inside the cavity is uniform and proportional to the distance between the center of the sphere and the center of the cavity ($\vec{a}$): $\vec{E} = \frac{\rho \vec{a}}{3\epsilon_0}$. This is a favorite "High Order Thinking" question in JEE Main.

Electric Potential & Potential Energy

The concept of potential is the scalar counterpart to the vector concept of the Electric Field. In JEE physics, mastering potential is crucial because scalar addition is significantly easier than vector addition, and it provides a direct bridge to the Conservation of Energy.

1. Electric Potential Energy ($U$)

Electric Potential Energy is defined as the work done by an external agent in assembling a system of charges by bringing them from infinity to their respective positions without any change in kinetic energy.

For two point charges $q_1$ and $q_2$ separated by a distance $r$: $$U = \frac{k q_1 q_2}{r} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$$

Key Characteristics:

1. Nature: It is a scalar quantity.
2. Unit: Joule (J) or electron-Volt (eV). $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.
3. Sign Dependency: $U$ is positive if the charges are of the same sign (repulsive force, work done against the field) and negative if they are of opposite signs (attractive force).

2. Electric Potential ($V$)

Electric Potential at a point is defined as the amount of work done per unit positive test charge in bringing it from infinity to that point.

$$V = \frac{W_{ext}}{q_0} = \frac{U}{q}$$

For a point charge $Q$ at a distance $r$: $$V = \frac{kQ}{r}$$

The Principle of Superposition

For a system of $n$ charges, the net potential at a point $P$ is the algebraic sum of individual potentials: $$V_{net} = \sum_{i=1}^{n} \frac{k q_i}{r_i} = V_1 + V_2 + V_3 + ...$$ Note: We must substitute the signs of the charges while calculating potential.

3. Potential Due to Continuous Charge Distributions

To find the potential of a continuous object, we use $V = \int \frac{k dq}{r}$.

A. Electric Dipole

The potential at point $P(r, \theta)$ where $r \gg l$: $$V = \frac{k p \cos \theta}{r^2} = \frac{k \vec{p} \cdot \hat{r}}{r^2}$$

• At Axial Point ($\theta = 0^\circ$): $V = \frac{kp}{r^2}$
• At Equatorial Point ($\theta = 90^\circ$): $V = 0$

B. Uniformly Charged Ring

For a ring of radius $R$ and total charge $Q$, at a point $P$ on its axis at distance $x$ from the center: $$V_{axis} = \frac{kQ}{\sqrt{R^2 + x^2}}$$

• At the center ($x=0$): $V_c = \frac{kQ}{R}$

C. Uniformly Charged Disc

For a disc of radius $R$ and surface charge density $\sigma$, at an axial point $x$: $$V = \frac{\sigma}{2\epsilon_0} \left[ \sqrt{R^2 + x^2} - x \right]$$

D. Spherical Shell (Conducting Sphere)

For a shell of radius $R$ and charge $Q$:

1. Outside ($r > R$): $V_{out} = \frac{kQ}{r}$
2. Surface ($r = R$): $V_s = \frac{kQ}{R}$
3. Inside ($r < R$): $V_{in} = \frac{kQ}{R}$ (Potential is constant throughout the volume)

E. Non-Conducting Uniformly Charged Solid Sphere

For a sphere of radius $R$ and total charge $Q$:

1. Outside ($r \ge R$): $V_{out} = \frac{kQ}{r}$
2. Inside ($r < R$): $V_{in} = \frac{kQ}{2R^3} (3R^2 - r^2)$
   • At the center ($r=0$): $V_{center} = \frac{3kQ}{2R} = 1.5 V_{surface}$

4. Relation between Electric Field and Potential

Electric field is the negative gradient of electric potential.

Vector Relation

$$\vec{E} = -\vec{\nabla} V = -\left[ \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right]$$ Where $E_x = -\frac{\partial V}{\partial x}$, $E_y = -\frac{\partial V}{\partial y}$, and $E_z = -\frac{\partial V}{\partial z}$.

Integral Relation

If the field is known, the potential difference $\Delta V$ between two points $A$ and $B$ is: $$V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r}$$

For a uniform electric field: $$V = E \times d$$ (where $d$ is the distance along the field line).

5. Equipotential Surfaces

An equipotential surface is the locus of all points in space which are at the same potential.

Geometry of Surfaces:

• Point Charge: Concentric Spheres.
• Uniform Field: Parallel Planes.
• Line Charge: Coaxial Cylinders.

6. Electrostatic Self-Energy

Self-energy is the energy required to build a charged distribution from scratch.

1. Uniformly Charged Conducting Sphere (or Shell): $$U_{self} = \frac{kQ^2}{2R}$$
2. Uniformly Charged Non-Conducting Solid Sphere: $$U_{self} = \frac{3kQ^2}{5R}$$

This energy represents the work done against growing repulsion as infinitesimal charges are added to the surface or volume.

7. The Work-Energy Theorem in Electrostatics

In electrostatics, the work done by the conservative electric force is $W_{elec} = -\Delta U$. By the Work-Energy Theorem: $$W_{ext} + W_{elec} = \Delta K$$

If the charge is moved slowly ($\Delta K = 0$): $$W_{ext} = \Delta U = q(V_f - V_i)$$

If a charge $q$ is accelerated through a potential difference $V$: $$qV = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

8. Worked Examples

Example 1: Potential from a System of Points

Three charges $+q, -2q$, and $+q$ are placed at vertices of an equilateral triangle of side $L$. Find the potential at the centroid. Solution:

1. Distance from vertex to centroid $r = \frac{L}{\sqrt{3}}$.
2. $V_{net} = V_1 + V_2 + V_3$.
3. $V_{net} = \frac{kq}{L/\sqrt{3}} + \frac{k(-2q)}{L/\sqrt{3}} + \frac{kq}{L/\sqrt{3}}$.
4. $V_{net} = \frac{\sqrt{3}k}{L} (q - 2q + q) = 0$. Answer: $0$.

Example 2: Field from Potential Function

If $V(x, y, z) = 3x^2y - y^3z$, find the electric field $\vec{E}$ at $(1, 2, 0)$. Solution:

1. $E_x = -\frac{\partial V}{\partial x} = -(6xy) = -6(1)(2) = -12$.
2. $E_y = -\frac{\partial V}{\partial y} = -(3x^2 - 3y^2z) = -(3(1)^2 - 0) = -3$.
3. $E_z = -\frac{\partial V}{\partial z} = -(-y^3) = y^3 = 2^3 = 8$.
4. $\vec{E} = -12\hat{i} - 3\hat{j} + 8\hat{k} \text{ V/m}$.

Example 3: Work Done in Moving Charge

Find the work done by an external agent to move a charge $q = 2\mu C$ from point $A(2, 0)$ to $B(0, 2)$ in a uniform field $\vec{E} = 100 \hat{i} \text{ V/m}$. Solution:

1. $V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r} = -\int (100 \hat{i}) \cdot (dx \hat{i} + dy \hat{j})$.
2. $V_B - V_A = -100 \int_2^0 dx = -100 [x]_2^0 = -100(0 - 2) = 200 \text{ V}$.
3. $W_{ext} = q(V_B - V_A) = (2 \times 10^{-6})(200) = 4 \times 10^{-4} \text{ J}$.

Example 4: Potential of a Charged Ring and Point Charge

A charge $+Q$ is fixed on a ring of radius $R$. A point charge $-q$ is placed at distance $x$ on the axis. Find the velocity of $-q$ when it reaches the center if released from rest. Solution:

1. Apply Energy Conservation: $U_i + K_i = U_f + K_f$.
2. $U_i = \frac{kQ(-q)}{\sqrt{R^2 + x^2}}$, $K_i = 0$.
3. $U_f = \frac{kQ(-q)}{R}$, $K_f = \frac{1}{2}mv^2$.
4. $-\frac{kQq}{\sqrt{R^2 + x^2}} = -\frac{kQq}{R} + \frac{1}{2}mv^2$.
5. $v = \sqrt{\frac{2kQq}{m} \left( \frac{1}{R} - \frac{1}{\sqrt{R^2 + x^2}} \right)}$.

Example 5: Concentric Shells

Two concentric conducting shells of radii $R_1$ and $R_2$ ($R_1 < R_2$) have charges $Q_1$ and $Q_2$. Find the potential of the inner shell. Solution:

1. Potential on inner shell $V_1 = V_{self} + V_{due_to_outer}$.
2. Inner shell surface is inside the outer shell.
3. $V_1 = \frac{kQ_1}{R_1} + \frac{kQ_2}{R_2}$.

Example 6: Self-Energy Comparison

If a solid non-conducting sphere and a hollow shell have the same charge $Q$ and radius $R$, find the ratio of their self-energies. Solution:

1. $U_{solid} = \frac{3kQ^2}{5R}$.
2. $U_{shell} = \frac{kQ^2}{2R}$.
3. Ratio $\frac{U_{solid}}{U_{shell}} = \frac{3/5}{1/2} = \frac{6}{5} = 1.2$.

Example 7: Dipole Potential at an Angle

A dipole of moment $p = 4 \times 10^{-9} \text{ Cm}$ is placed at the origin. Find the potential at a point $r=2\text{m}$ making $60^\circ$ with the dipole axis. Solution:

1. $V = \frac{kp \cos \theta}{r^2}$.
2. $V = \frac{(9 \times 10^9)(4 \times 10^{-9}) \cos 60^\circ}{2^2} = \frac{36 \times 0.5}{4} = 4.5 \text{ V}$.

Example 8: Equipotential of Line Charge

The potential due to an infinite line charge is $V_2 - V_1 = 2k\lambda \ln(r_1/r_2)$. If $V=0$ is taken at $r=r_0$, find the potential at $r$. Solution:

1. $V_r - V_{r_0} = - \int_{r_0}^r \frac{2k\lambda}{r} dr$.
2. $V_r = 2k\lambda \ln(r_0/r)$.

Example 9: Assembling a Square

Four charges $+q$ are placed at the corners of a square of side $a$. Find the total potential energy of the system. Solution:

1. Total pairs = $_4C_2 = 6$.
2. 4 pairs are sides (distance $a$), 2 pairs are diagonals (distance $a\sqrt{2}$).
3. $U = 4 \left( \frac{kq^2}{a} \right) + 2 \left( \frac{kq^2}{a\sqrt{2}} \right) = \frac{kq^2}{a} [4 + \sqrt{2}]$.

Example 10: Droplet Coalescence

$N$ identical mercury droplets, each at potential $V_0$, coalesce to form a single large drop. Find the potential of the large drop. Solution:

1. Volume conservation: $\frac{4}{3} \pi R^3 = N \times \frac{4}{3} \pi r^3 \implies R = N^{1/3} r$.
2. Charge conservation: $Q = Nq$.
3. $V_{new} = \frac{kQ}{R} = \frac{k(Nq)}{N^{1/3}r} = N^{2/3} \frac{kq}{r} = N^{2/3} V_0$.

Mnemonics for Revision

• For Shells/Spheres: "Inside is safe, outside is point." — Inside a shell, potential is constant (safe from changes); outside, it behaves like a point charge at the center.
• Field-Potential Relation: "Electric field points to lower potential." — Just like water flows from high pressure to low, $\vec{E}$ points down the potential hill.
• Self Energy Solid: $3/5$ is "three-fifths of a star" (Solid stars are dense).

Capacitance & Capacitors

Capacitance is the ability of a system to store electric charge and, consequently, electrostatic potential energy. In competitive exams like JEE Main, this chapter bridges the gap between static electrostatics and dynamic current electricity.

1. Fundamental Definition of Capacitance

When a conductor is given a charge $Q$, its potential $V$ increases proportionally. $$Q \propto V \implies Q = CV$$ Where $C$ is the Capacitance.

• SI Unit: Farad (F). $1\text{ F} = 1\text{ Coulomb/Volt}$.
• Nature: It is a geometric property. It depends on the shape, size, and medium; it does not depend on $Q$ or $V$ individually.

1.1 Capacitance of an Isolated Spherical Conductor

For a sphere of radius $R$ in vacuum, the potential at the surface is $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$. $$C = \frac{Q}{V} = 4\pi\epsilon_0 R$$ Mnemonic: If $R = 6400\text{ km}$ (Earth), $C \approx 711\text{ \mu F}$. This shows that 1 Farad is an enormous unit.

2. Standard Capacitor Geometries

A capacitor usually consists of two conductors with charges $+Q$ and $-Q$. The capacitance is defined as $C = \frac{Q}{\Delta V}$.

2.1 The Parallel Plate Capacitor (PPC)

The electric field between the plates is $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$. The potential difference is $V = Ed = \frac{Qd}{A\epsilon_0}$. Thus, $C = \frac{\epsilon_0 A}{d}$.

3. Energy Stored in a Capacitor

The process of charging a capacitor involves doing work against the electrostatic repulsion. This work is stored as Potential Energy ($U$). $$U = \int V dQ = \int \frac{Q}{C} dQ = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2 = \frac{1}{2}QV$$

3.1 Energy Density ($u$)

Energy is stored in the electric field. Energy density is the energy per unit volume: $$u = \frac{U}{\text{Volume}} = \frac{\frac{1}{2}CV^2}{Ad} = \frac{\frac{1}{2}(\frac{\epsilon_0 A}{d})(Ed)^2}{Ad}$$ $$u = \frac{1}{2}\epsilon_0 E^2$$

4. Dielectrics in Capacitors

A dielectric is an insulator that polarizes when placed in an electric field. This polarization creates "bound charges" on the surface, which produce an induced field ($E_{ind}$) opposing the external field ($E_0$).

• Dielectric Constant ($K$): The ratio by which the field is reduced. $E_{net} = \frac{E_0}{K}$.
• New Capacitance: $C = \frac{K\epsilon_0 A}{d} = KC_0$.

4.1 Induced Bound Charges

The induced charge density $\sigma_b$ is given by: $$\sigma_b = \sigma \left( 1 - \frac{1}{K} \right)$$ Where $\sigma$ is the free charge density on the metal plates.

4.2 Force between Plates

Since the plates have opposite charges, they attract each other. $$F = \frac{Q^2}{2A\epsilon_0} = \frac{1}{2}\epsilon_0 E^2 A$$ Note: Use the field of one plate ($E/2$) to calculate force on the other.

5. Combination of Capacitors

6. Insertion of Dielectric Slabs: Two Cases

JEE often asks about the change in $Q, V, E, C, U$ when a slab is inserted.

7. Redistribution of Charge and Energy Loss

When two capacitors $C_1$ (at $V_1$) and $C_2$ (at $V_2$) are connected in parallel, charges flow until they reach a Common Potential ($V_c$).

$$V_c = \frac{Q_{total}}{C_{total}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$

During this flow, energy is dissipated as heat/EM radiation ($\Delta H$). $$\Delta H = U_i - U_f = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$$

8. 15 Solved JEE-Style Problems

Problem 1: Basic PPC Calculation Calculate the capacitance of a parallel plate capacitor with plate area $20\text{ cm}^2$ separated by $2\text{ mm}$ of air. Solution: $A = 20 \times 10^{-4}\text{ m}^2$, $d = 2 \times 10^{-3}\text{ m}$, $\epsilon_0 = 8.85 \times 10^{-12}$. $C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 20 \times 10^{-4}}{2 \times 10^{-3}} = 8.85 \times 10^{-12}\text{ F} = 8.85\text{ pF}$.

Problem 2: Energy Stored A $10\text{ \mu F}$ capacitor is charged to $50\text{ V}$. Find the energy stored. Solution: $U = \frac{1}{2}CV^2 = \frac{1}{2} (10 \times 10^{-6}) (50)^2 = 5 \times 10^{-6} \times 2500 = 0.0125\text{ J} = 12.5\text{ mJ}$.

Problem 3: Dielectric Effect A capacitor has capacitance $C$. A dielectric of $K=4$ fills half the gap thickness-wise. Find new capacitance. Solution: This is a series combination of two capacitors: $C_1$ with air ($d/2$) and $C_2$ with dielectric ($d/2$). $C_1 = \frac{\epsilon_0 A}{d/2} = 2C_0$; $C_2 = \frac{K\epsilon_0 A}{d/2} = 2KC_0 = 8C_0$. $1/C_{eq} = 1/2C_0 + 1/8C_0 = 5/8C_0 \implies C_{eq} = 1.6C_0$.

Problem 4: Series-Parallel Mixing Three $6\text{ \mu F}$ capacitors are connected in series. This combination is in parallel with a $2\text{ \mu F}$ capacitor. Final $C_{eq}$? Solution: Series part: $1/C_s = 1/6 + 1/6 + 1/6 = 3/6 \implies C_s = 2\text{ \mu F}$. Parallel part: $C_{eq} = C_s + 2 = 2 + 2 = 4\text{ \mu F}$.

Problem 5: Capacitor with Variable Dielectric A PPC has a dielectric slab that varies as $K(x) = K_0(1 + \frac{x}{d})$ where $x$ is distance from one plate. Area is $A$. Find $C$. Solution: Take a small element $dx$. $dC = \frac{K(x)\epsilon_0 A}{dx}$. These elements are in series: $\frac{1}{C} = \int_{0}^{d} \frac{1}{dC} = \int_{0}^{d} \frac{dx}{K_0(1 + x/d)\epsilon_0 A}$. $\frac{1}{C} = \frac{1}{K_0 \epsilon_0 A} [d \ln(1 + x/d)]_0^d = \frac{d \ln 2}{K_0 \epsilon_0 A}$. $C = \frac{K_0 \epsilon_0 A}{d \ln 2}$.

Problem 6: Heat Loss in Redistribution $C_1 = 2\text{ \mu F}$ at $10\text{ V}$ is connected to $C_2 = 3\text{ \mu F}$ at $20\text{ V}$ (opposite polarities). Find common potential. Solution: Opposite polarities: $Q_{net} = |C_1V_1 - C_2V_2| = |2(10) - 3(20)| = |20 - 60| = 40\text{ \mu C}$. $V_c = \frac{Q_{net}}{C_1 + C_2} = \frac{40}{2+3} = 8\text{ V}$.

Problem 7: Force on Plate Charge $Q = 10\text{ \mu C}$, Area $A = 1\text{ m}^2$. Find force between plates. Solution: $F = \frac{Q^2}{2A\epsilon_0} = \frac{(10 \times 10^{-6})^2}{2 \times 1 \times 8.85 \times 10^{-12}} = \frac{10^{-10}}{17.7 \times 10^{-12}} = \frac{100}{17.7} \approx 5.65\text{ N}$.

Problem 8: Spherical Capacitor Inner radius $1\text{ cm}$, Outer radius $2\text{ cm}$. Vacuum. Find $C$. Solution: $C = \frac{4\pi\epsilon_0 ab}{b-a} = \frac{1}{9\times 10^9} \frac{10^{-2} \times 2 \times 10^{-2}}{10^{-2}} = \frac{2\times 10^{-2}}{9\times 10^9} \approx 2.22\text{ pF}$.

Problem 9: Cylindrical Capacitor Length $2\text{ m}$, $a=1\text{ mm}$, $b=2\text{ mm}$. Find $C$. Solution: $C = \frac{2\pi\epsilon_0 l}{\ln(b/a)} = \frac{l}{2k \ln(2)} = \frac{2}{2 \times 9\times 10^9 \times 0.693} \approx 160\text{ pF}$.

Problem 10: Energy Stored in Electric Field A field of $10^6\text{ V/m}$ exists in a volume of $1\text{ cm}^3$. Find energy. Solution: $u = \frac{1}{2}\epsilon_0 E^2 = 0.5 \times 8.85 \times 10^{-12} \times 10^{12} = 4.425\text{ J/m}^3$. Energy $U = u \times \text{Vol} = 4.425 \times 10^{-6}\text{ J} = 4.425\text{ \mu J}$.

Problem 11: Breaking Voltage Three capacitors of $1\text{ \mu F}$ each with max voltage $10\text{ V}$ are in series. Max voltage of combination? Solution: In series, $1/C_{eq} = 3 \implies C_{eq} = 1/3\text{ \mu F}$. Since capacitors are identical, $V$ divides equally. So $V_{max} = 3 \times 10\text{ V} = 30\text{ V}$.

Problem 12: Dielectric Slab Insertion (Battery Removed) $C = 10\text{ \mu F}$, $V = 100\text{ V}$. Battery removed. Dielectric $K=5$ inserted. New Energy? Solution: $U_0 = \frac{1}{2}CV^2 = 0.5 \times 10\text{ \mu F} \times 100^2 = 0.05\text{ J}$. With battery removed, $U_{new} = U_0 / K = 0.05 / 5 = 0.01\text{ J}$.

Problem 13: Dielectric Slab Insertion (Battery Connected) Same as above, but battery remains. New Energy? Solution: $U_{new} = K U_0 = 5 \times 0.05 = 0.25\text{ J}$.

Problem 14: Partial Dielectric (Area-wise) A PPC has half the area filled with $K_1$ and half with $K_2$. Find $C$. Solution: These are in parallel. $C_1 = \frac{K_1 \epsilon_0 (A/2)}{d}$, $C_2 = \frac{K_2 \epsilon_0 (A/2)}{d}$. $C_{eq} = C_1 + C_2 = \frac{\epsilon_0 A}{2d} (K_1 + K_2)$.

Problem 15: Charge on one plate In a $5\text{ \mu F}$ capacitor, the potential difference is $20\text{ V}$. What is the charge on the negative plate? Solution: $Q = CV = 5 \times 10^{-6} \times 20 = 100\text{ \mu C}$. Charge on positive plate $= +100\text{ \mu C}$, Charge on negative plate $= -100\text{ \mu C}$.

9. Key Exam Strategies

• Infinite Ladders: Use the concept that adding one more unit to an infinite chain doesn't change $C_{eq}$.
• Wheatstone Bridge: If $\frac{C_1}{C_2} = \frac{C_3}{C_4}$, the central capacitor has no charge.
• Circuit Symmetry: Use folding or mirror symmetry to simplify complex grids.

Current, Drift Velocity & Ohm's Law

In the realm of Electrostatics, we dealt with charges at rest. However, the heart of modern technology lies in Electrodynamics—the study of charges in motion. This section explores the fundamental mechanisms of charge transport in conductors, establishing the link between microscopic electron motion and macroscopic electrical parameters.

1. Electric Current ($I$) and Current Density ($J$)

Electric Current ($I$) is defined as the rate of flow of charge across any cross-section of a conductor. If a net charge $dq$ passes through a cross-section in time $dt$, the instantaneous current is: $$I = \frac{dq}{dt}$$

• Unit: Ampere ($A$) where $1 A = 1 C/s$.
• Nature: It is a scalar quantity. Although we assign a direction to current (conventional current flows from higher to lower potential), it does not follow vector addition laws.
• Steady Current: If the flow of charge is constant over time, $I = \frac{Q}{t}$.

Current Density ($\vec{J}$) is a vector quantity that describes the flow of charge at a specific point. It is defined as the current per unit area of cross-section held perpendicular to the direction of flow. $$\vec{J} = \frac{dI}{dA} \hat{n} \implies I = \int \vec{J} \cdot d\vec{A}$$ If the current is uniform and the area $A$ is perpendicular to flow: $$J = \frac{I}{A}$$

• Unit: $A/m^2$.
• Direction: Same as the direction of motion of positive charges.

2. The Mechanism of Drift Velocity ($v_d$)

In a conductor, free electrons move randomly due to thermal energy (thermal speeds $\approx 10^5$ to $10^6$ m/s). Because this motion is random, the average velocity is zero, and no net current flows.

When an external electric field $\vec{E}$ is applied, each electron experiences an electric force $\vec{F} = -e\vec{E}$, resulting in an acceleration: $$\vec{a} = \frac{-e\vec{E}}{m}$$ where $m$ is the mass of an electron.

The electrons frequently collide with the heavy, fixed ions of the lattice. After each collision, the electron loses its gained momentum and starts accelerating again. The average time interval between two successive collisions is called the Relaxation Time ($\tau$).

Derivation of the Expression for $v_d$

Let $u_i$ be the initial thermal velocity of the $i^{th}$ electron. After time $t_i$ (the time since its last collision), its velocity $v_i$ is: $$v_i = u_i + a t_i$$ The average velocity of all $n$ electrons is the drift velocity $v_d$: $$v_d = \text{avg}(v_i) = \frac{\sum u_i}{n} + a \frac{\sum t_i}{n}$$ Since the average thermal velocity $\frac{\sum u_i}{n} = 0$, and the average time between collisions is $\tau$: $$v_d = 0 + a\tau$$ Substituting $a = \frac{eE}{m}$ (considering magnitude): $$v_d = \frac{eE\tau}{m}$$

Relation between $I$ and $v_d$

Consider a conductor of length $l$ and cross-sectional area $A$. Let $n$ be the number of free electrons per unit volume (number density). Total number of electrons in volume $Al = nAl$. Total charge $Q = (nAl)e$. Time taken for these charges to cross the area is $t = \frac{l}{v_d}$. Current $I = \frac{Q}{t} = \frac{nAle}{l/v_d}$ $$I = n e A v_d$$ Therefore, current density $J = \frac{I}{A} = n e v_d$.

3. Mobility ($\mu$)

Mobility is defined as the magnitude of drift velocity per unit electric field. It tells us how "mobile" a charge carrier is within a medium. $$\mu = \frac{v_d}{E} = \frac{e\tau}{m}$$

• Unit: $m^2 V^{-1} s^{-1}$.
• Mobility is always positive regardless of the charge of the carrier.

4. Microscopic and Macroscopic Ohm's Law

Microscopic Form

From $J = n e v_d$ and $v_d = \frac{eE\tau}{m}$, we get: $$J = n e \left( \frac{eE\tau}{m} \right) = \left( \frac{n e^2 \tau}{m} \right) E$$ The term in the bracket is a constant for a material at a given temperature, called Conductivity ($\sigma$). $$\sigma = \frac{n e^2 \tau}{m}$$ Thus, $\vec{J} = \sigma \vec{E}$. This is the Microscopic Ohm's Law.

Macroscopic Form

We know $V = El$ and $I = JA$. Substituting $J = \sigma E$: $$\frac{I}{A} = \sigma \left( \frac{V}{l} \right) \implies V = I \left( \frac{l}{\sigma A} \right)$$ Define Resistivity ($\rho$) as $\rho = \frac{1}{\sigma} = \frac{m}{n e^2 \tau}$. Then, $V = I \left( \frac{\rho l}{A} \right)$. Defining $R = \frac{\rho l}{A}$ as the Resistance, we get: $$V = IR$$ This is the standard Ohm's Law (Macroscopic form), valid under constant physical conditions (like temperature).

5. Temperature Dependence of Resistivity

The resistivity $\rho = \frac{m}{ne^2\tau}$ depends on two variables that change with temperature:

1. $n$ (carrier concentration): Increases with temperature (significant in semiconductors).
2. $\tau$ (relaxation time): Decreases as temperature increases due to more frequent lattice vibrations/collisions.

Temperature Coefficient of Resistance ($\alpha$): For small temperature ranges: $$R_t = R_0 [1 + \alpha(T - T_0)]$$ $$\rho_t = \rho_0 [1 + \alpha(T - T_0)]$$

• For metals: $\alpha > 0$ (Positive).
• For semiconductors/insulators: $\alpha < 0$ (Negative).

6. Resistor Colour Coding

Carbon resistors are coded with four colour bands to indicate their value.

The Mnemonic: "BB ROY of Great Britain had a Very Good Wife."

Formula: $R = (AB \times 10^C \pm D) \Omega$ Where A=1st digit, B=2nd digit, C=Multiplier, D=Tolerance.

7. Combinations of Resistors

Series Combination

• Current is same through all resistors.
• $V_{total} = V_1 + V_2 + \dots$
• $R_{eq} = R_1 + R_2 + R_3 + \dots$
• $R_{eq}$ is always greater than the largest individual resistance.

Parallel Combination

• Potential Difference ($V$) is same across all resistors.
• $I_{total} = I_1 + I_2 + \dots$
• $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$
• For two resistors: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$
• $R_{eq}$ is always smaller than the smallest individual resistance.

8. Solved Problems (JEE Level)

Problem 1: Drift Velocity Calculation A copper wire of cross-sectional area $10^{-7} m^2$ carries a current of $1.5 A$. If the number density of free electrons is $8 \times 10^{28} m^{-3}$, calculate the drift velocity. Solution: Using $I = neAv_d$: $$v_d = \frac{I}{neA}$$ $$v_d = \frac{1.5}{(8 \times 10^{28}) \times (1.6 \times 10^{-19}) \times 10^{-7}}$$ $$v_d = \frac{1.5}{12.8 \times 10^2} \approx 1.17 \times 10^{-3} m/s$$ Ans: $1.17 mm/s$.

Problem 2: Stretched Wire Resistance A wire of resistance $R$ is stretched to double its original length. Find the new resistance assuming density and resistivity remain constant. Solution: Original $R = \rho \frac{l}{A}$. When stretched, volume $V = Al$ remains constant. New length $l' = 2l$. Since $A'l' = Al \implies A'(2l) = Al \implies A' = A/2$. New Resistance $R' = \rho \frac{l'}{A'}$ $$R' = \rho \frac{2l}{A/2} = 4 \left( \rho \frac{l}{A} \right) = 4R$$ Note: For a wire stretched $n$ times, $R_{new} = n^2 R_{old}$.

Problem 3: Microscopic Ohm's Law Application Determining the relaxation time $\tau$ for a metal with $n = 10^{29} m^{-3}$ and resistivity $\rho = 1.7 \times 10^{-8} \Omega m$. ($m_e = 9.1 \times 10^{-31} kg$) Solution: We use $\rho = \frac{m}{ne^2\tau} \implies \tau = \frac{m}{ne^2\rho}$ $$\tau = \frac{9.1 \times 10^{-31}}{10^{29} \times (1.6 \times 10^{-19})^2 \times 1.7 \times 10^{-8}}$$ $$\tau = \frac{9.1 \times 10^{-31}}{10^{29} \times 2.56 \times 10^{-38} \times 1.7 \times 10^{-8}}$$ $$\tau \approx \frac{9.1 \times 10^{-31}}{4.35 \times 10^{-17}} \approx 2.1 \times 10^{-14} s$$

Problem 4: Temperature Coefficient The resistance of a conductor is $5 \Omega$ at $50^\circ C$ and $6 \Omega$ at $100^\circ C$. Find the resistance at $0^\circ C$. Solution: Using $R_t = R_0(1 + \alpha \Delta T)$: $5 = R_0(1 + 50\alpha)$ --- (i) $6 = R_0(1 + 100\alpha)$ --- (ii) Dividing (ii) by (i): $\frac{6}{5} = \frac{1 + 100\alpha}{1 + 50\alpha} \implies 6 + 300\alpha = 5 + 500\alpha$ $1 = 200\alpha \implies \alpha = 0.005 /^\circ C$. Substitute $\alpha$ in (i): $5 = R_0(1 + 50 \times 0.005) = R_0(1 + 0.25)$ $5 = 1.25 R_0 \implies R_0 = \frac{5}{1.25} = 4 \Omega$. Ans: $4 \Omega$.

Problem 5: Current Density in Non-uniform Wire A current of $5A$ flows through a wire whose area changes linearly from $1 mm^2$ to $2 mm^2$ over a length $L$. Find the current density at the midpoint. Solution: Midpoint area for a linear variation $A_{mid} = \frac{A_1 + A_2}{2}$. $A_{mid} = \frac{1 + 2}{2} = 1.5 mm^2 = 1.5 \times 10^{-6} m^2$. $J = \frac{I}{A} = \frac{5}{1.5 \times 10^{-6}} = 3.33 \times 10^6 A/m^2$.

Problem 6: Colour Code Interpretation A resistor has bands of Yellow, Violet, Orange, and Silver. Calculate its resistance range. Solution: Yellow = 4, Violet = 7, Orange = $10^3$, Silver = 10%. $R = 47 \times 10^3 \Omega \pm 10% = 47 k\Omega \pm 4.7 k\Omega$. Range: $42.3 k\Omega$ to $51.7 k\Omega$.

Problem 7: Infinite Ladder Network Find the equivalent resistance between A and B for an infinite ladder of $1\Omega$ resistors. (Each horizontal and vertical branch is $1\Omega$). Solution: Let the total resistance be $X$. Adding one more "unit" to an infinite ladder doesn't change $X$. The unit consists of $1\Omega$ in series with the parallel combination of $1\Omega$ and $X$. $X = 1 + \left( \frac{1 \cdot X}{1 + X} \right)$ $X = \frac{1+X+X}{1+X} \implies X(1+X) = 1+2X$ $X + X^2 = 1 + 2X \implies X^2 - X - 1 = 0$ Using quadratic formula: $X = \frac{1 + \sqrt{1 + 4}}{2} = \frac{1 + \sqrt{5}}{2} \Omega$ (Taking positive root). Ans: $\approx 1.618 \Omega$.

Problem 8: Mobility of Charge Carriers In a semiconductor, the concentration of electrons is $8 \times 10^{19} m^{-3}$ and hole concentration is $5 \times 10^{18} m^{-3}$. If mobility of $e^-$ is $2.3 m^2 V^{-1} s^{-1}$ and holes is $0.01 m^2 V^{-1} s^{-1}$, find the conductivity. Solution: Total $\sigma = \sigma_e + \sigma_h = n_e e \mu_e + n_h e \mu_h$ $\sigma = e (n_e \mu_e + n_h \mu_h)$ $\sigma = 1.6 \times 10^{-19} [ (8 \times 10^{19} \times 2.3) + (5 \times 10^{18} \times 0.01) ]$ $\sigma = 1.6 \times 10^{-19} [ 18.4 \times 10^{19} + 0.05 \times 10^{18} ]$ $\sigma \approx 1.6 \times 10^{-19} \times 18.4 \times 10^{19} = 29.44 \Omega^{-1} m^{-1}$.

EMF, Internal Resistance & Kirchhoff's Laws

In the realm of Electrostatics, we deal with charges at rest. However, in Current Electricity, we focus on the dynamics of charges driven by a potential difference. To maintain a steady flow of current through a conductor, we need a source that performs work on the charge carriers to move them from a lower potential to a higher potential. This section explores the energetics of circuits and the fundamental laws governing complex network analysis.

1. Electromotive Force (EMF) and Internal Resistance

1.1 Electromotive Force (EMF)

Despite the name, EMF is not a force. It is a measure of energy per unit charge.

• Definition: The EMF ($\epsilon$) of a source is the work done by the source in taking a unit positive charge from the lower to the higher potential terminal when the circuit is open.
• Unit: Volt ($1 \text{ V} = 1 \text{ J/C}$).

1.2 Internal Resistance ($r$)

No cell is "ideal." The electrolytes and electrodes within a cell offer resistance to the flow of current. This inherent resistance is called Internal Resistance. It depends on:

1. Concentration of the electrolyte.
2. Distance between electrodes.
3. Area of electrodes immersed in the electrolyte.
4. Temperature (decreases as temperature increases).

1.3 Terminal Voltage ($V$)

The potential difference across the terminals of a cell when current is flowing through the circuit is called Terminal Voltage.

Relation between $\epsilon$, $V$, and $r$:

1. Discharging (Current flows out of positive terminal): $$V = \epsilon - Ir$$ Here, $V < \epsilon$.
2. Charging (Current enters positive terminal): $$V = \epsilon + Ir$$ Here, $V > \epsilon$.
3. Open Circuit ($I=0$): $$V = \epsilon$$

2. Combination of Cells

In JEE problems, we often encounter networks of multiple cells. These can be simplified into a single equivalent cell ($\epsilon_{eq}, r_{eq}$).

2.1 Cells in Series

If $n$ cells are connected in series such that the positive terminal of one is connected to the negative of the next:

• $\epsilon_{eq} = \epsilon_1 + \epsilon_2 + ... + \epsilon_n$
• $r_{eq} = r_1 + r_2 + ... + r_n$
• If one cell is reversed: $\epsilon_{eq} = \epsilon_1 - \epsilon_2$.

2.2 Cells in Parallel

If $n$ cells are connected in parallel:

• Equivalent EMF: $$\epsilon_{eq} = \frac{\sum \frac{\epsilon_i}{r_i}}{\sum \frac{1}{r_i}} = \frac{\frac{\epsilon_1}{r_1} + \frac{\epsilon_2}{r_2} + ...}{\frac{1}{r_1} + \frac{1}{r_2} + ...}$$
• Equivalent Internal Resistance: $$\frac{1}{r_{eq}} = \sum \frac{1}{r_i}$$
• Special Case: If $n$ identical cells ($\epsilon, r$) are in parallel: $\epsilon_{eq} = \epsilon$ and $r_{eq} = r/n$.

2.3 Mixed Grouping

Consider $n$ identical cells in series across $m$ such rows in parallel, connected to an external resistor $R$.

• Total number of cells $N = nm$.
• Total EMF = $n\epsilon$.
• Total internal resistance $r_{total} = \frac{nr}{m}$.
• Circuit Current $I = \frac{n\epsilon}{R + \frac{nr}{m}} = \frac{mn\epsilon}{mR + nr}$.

3. Maximum Power Transfer Theorem

For a network with a constant EMF source $\epsilon$ and internal resistance $r$, the power delivered to an external load $R$ is: $$P = I^2 R = \left( \frac{\epsilon}{R+r} \right)^2 R$$

To find the condition for maximum power, we differentiate $P$ with respect to $R$ and set it to zero ($\frac{dP}{dR} = 0$):

• Condition: $R = r$ (External resistance = Internal resistance).
• Maximum Power: $P_{max} = \frac{\epsilon^2}{4r}$.
• Efficiency at Max Power: $\eta = 50%$.

4. Kirchhoff’s Laws

Standard Ohm’s Law is insufficient for multi-loop circuits. Kirchhoff's laws provide a systematic approach.

4.1 Kirchhoff’s First Law: Junction Rule (KCL)

• Statement: The algebraic sum of currents meeting at any junction in a circuit is zero.
• Conservation Principle: Law of Conservation of Charge.
• Equation: $\sum I_{in} = \sum I_{out}$.

4.2 Kirchhoff’s Second Law: Loop Rule (KVL)

• Statement: The algebraic sum of changes in potential around any closed loop involves resistors and cells is zero.
• Conservation Principle: Law of Conservation of Energy.
• Equation: $\sum \Delta V = 0$.

4.3 Sign Conventions (The "Universal" Method)

1. Resistors: If you move in the direction of current, $\Delta V = -IR$. If you move against the current, $\Delta V = +IR$.
2. Cells: If you move from negative to positive terminal, $\Delta V = +\epsilon$. If you move from positive to negative, $\Delta V = -\epsilon$.

5. Complex Circuit Analysis: Mesh & Nodal Methods

For JEE Advanced and tricky JEE Main problems, the Nodal Analysis (based on KCL) is usually faster than the Loop Method (KVL).

• Step 1: Select a reference node and set its potential to $0 \text{ V}$.
• Step 2: Assign potentials $x, y, z$ to other nodes.
• Step 3: Apply KCL at each unknown node: $\sum \frac{V_{node} - V_{adjacent}}{R} = 0$.
• Step 4: Solve the resulting linear equations.

6. Solved Multi-Loop Problems

Problem 1: Basic Two-Loop KVL

Find the current in each branch of a circuit with two loops where Loop 1 has $\epsilon_1=10\text{V}, r_1=2\Omega$ and Loop 2 has $\epsilon_2=4\text{V}, r_2=1\Omega$, sharing a middle resistor $R=5\Omega$.

Solution:

1. Let current $I_1$ flow from $\epsilon_1$ and $I_2$ from $\epsilon_2$. The middle branch carries $I_1 + I_2$.
2. Loop 1: $10 - 2I_1 - 5(I_1 + I_2) = 0 \Rightarrow 7I_1 + 5I_2 = 10$.
3. Loop 2: $4 - 1I_2 - 5(I_1 + I_2) = 0 \Rightarrow 5I_1 + 6I_2 = 4$.
4. Solving simultaneously: $I_1 = \frac{40}{17}\text{ A}$, $I_2 = -1.29\text{ A}$. The negative sign indicates $I_2$ flows opposite to the assumed direction.

Problem 2: The Symmetrical Bridge

Calculate the equivalent resistance of a Wheatstone bridge where all four outer resistors are $10\Omega$ and the galvanometer branch is $5\Omega$.

Solution:

1. Since $\frac{R_1}{R_2} = \frac{R_3}{R_4}$ ($10/10 = 10/10$), the bridge is balanced.
2. In a balanced bridge, the potential difference across the bridge (galvanometer) is zero.
3. No current flows through the $5\Omega$ resistor.
4. Remove the $5\Omega$ resistor. Now, the top two $10\Omega$ are in series ($20\Omega$), bottom two $10\Omega$ are in series ($20\Omega$).
5. $R_{eq} = 20 || 20 = 10\Omega$.

Problem 3: Unbalanced Wheatstone Bridge (Nodal Analysis)

Find $R_{eq}$ if $R_1=1, R_2=2, R_3=2, R_4=1$ and the middle arm is $1\Omega$.

Solution:

1. Assume a $1\text{V}$ battery across the bridge. Total current $I = I_{total}$. $R_{eq} = 1/I$.
2. Set Bottom node $= 0\text{V}$, Top node $= 1\text{V}$. Let bridge nodes be $x$ and $y$.
3. KCL at $x$: $\frac{x-1}{1} + \frac{x-0}{2} + \frac{x-y}{1} = 0 \Rightarrow 2.5x - y = 1$ (Eq 1).
4. KCL at $y$: $\frac{y-1}{2} + \frac{y-0}{1} + \frac{y-x}{1} = 0 \Rightarrow 2.5y - x = 0.5$ (Eq 2).
5. Solve: $x = 0.57\text{V}, y = 0.43\text{V}$.
6. $I_{total}$ entering at node 1: $\frac{1-x}{1} + \frac{1-y}{2} = (1-0.57) + \frac{1-0.43}{2} = 0.43 + 0.285 = 0.715\text{A}$.
7. $R_{eq} = 1/0.715 \approx 1.4\Omega$.

Problem 4: Infinite Ladder Network

Calculate $R_{eq}$ between terminals A and B for an infinite chain of resistors each of $1\Omega$.

Solution:

1. Let the total resistance be $X$.
2. If we add one more section (one series, one parallel), the resistance remains $X$ because $\infty + 1 = \infty$.
3. Equation: $R + \frac{R \cdot X}{R + X} = X$.
4. For $R=1$: $1 + \frac{X}{1+X} = X \Rightarrow 1+X+X = X+X^2 \Rightarrow X^2 - X - 1 = 0$.
5. Solving via quadratic formula: $X = \frac{1 + \sqrt{1 - 4(1)(-1)}}{2} = \frac{1+\sqrt{5}}{2} \approx 1.618\Omega$.

Problem 5: Cube Resistance (Skeleton Cube)

12 resistors of resistance $R$ each are connected along the edges of a cube. Find $R_{eq}$ across body diagonal.

Solution:

1. Use the Path Symmetry method. Connect a battery of $6\text{V}$ (for easy math).
2. At the starting corner, current splits into 3 edges. Due to symmetry, $I_1 = I_2 = I_3 = I/3$.
3. Calculate potentials: $V_{start} - V_{end} = (\frac{I}{3}R) + (\frac{I}{6}R) + (\frac{I}{3}R) = \frac{5}{6}IR$.
4. $R_{eq} = \frac{V}{I} = \frac{5}{6}R$.

Problem 6: Mixed Grouping for Max Power

We have 36 cells, each $\epsilon=2\text{V}, r=0.5\Omega$. What combination gives max power to $R=2\Omega$?

Solution:

1. $nm = 36$.
2. Condition for max power: $R = \frac{nr}{m} \Rightarrow 2 = \frac{n(0.5)}{m} \Rightarrow 4m = n$.
3. Substitute $n$: $(4m)m = 36 \Rightarrow m^2 = 9 \Rightarrow m = 3$.
4. Then $n = 36/3 = 12$.
5. Answer: 12 cells in series, 3 such rows in parallel.

Problem 7: Potentiometer Balancing

A cell of $\epsilon=1.5\text{V}$ balances at $60\text{cm}$ on a potentiometer wire. Another cell balances at $80\text{cm}$. Find $\epsilon_2$.

Solution:

1. Principle: $\epsilon \propto l$.
2. $\frac{\epsilon_1}{\epsilon_2} = \frac{l_1}{l_2} \Rightarrow \frac{1.5}{\epsilon_2} = \frac{60}{80}$.
3. $\epsilon_2 = \frac{1.5 \times 80}{60} = 2.0\text{V}$.

Problem 8: Internal Resistance Calculation

A cell balances at $3\text{m}$ in open circuit. When $R=10\Omega$ is connected across it, balance point shifts to $2.5\text{m}$. Find $r$.

Solution:

1. Formula: $r = R \left( \frac{l_1}{l_2} - 1 \right)$.
2. $r = 10 \left( \frac{3}{2.5} - 1 \right) = 10(1.2 - 1) = 2\Omega$.

Problem 9: Superposition in Circuits

A circuit has two batteries ($20\text{V}$ and $10\text{V}$) and three resistors. Find current using superposition.

Solution:

1. Short the $10\text{V}$ source, calculate $I'_1$ due to $20\text{V}$.
2. Short the $20\text{V}$ source, calculate $I''_1$ due to $10\text{V}$.
3. $I_{net} = I'_1 + I''_1$ (be careful with directions).

Problem 10: Star-Delta Transformation

Convert a delta network of three $3\Omega$ resistors to a star network.

Solution:

1. Formula for $R_{star} = \frac{R_{adj1} R_{adj2}}{\sum R_{delta}}$.
2. $R_A = \frac{3 \times 3}{3+3+3} = \frac{9}{9} = 1\Omega$.
3. By symmetry, all star resistors are $1\Omega$.

7. Mnemonics and Quick Tips

Calculation Hack for JEE: In most multi-loop resistor problems, look for symmetry first. If the circuit can be folded or rotated such that it looks identical, the nodes on the axis of symmetry will be at the same potential (Equipotential nodes). You can merge these nodes to simplify the circuit significantly before applying Kirchhoff's Laws.

Wheatstone Bridge, Meter Bridge & Potentiometer

In the realm of Current Electricity, the measurement of unknown resistance and electromotive force (EMF) is of paramount importance. While a simple Voltmeter-Ammeter method can give approximate values, the precision required for JEE Main necessitates the use of Null-Deflection Instruments. This section covers the theoretical and practical aspects of the Wheatstone Bridge and its derivatives.

1. The Wheatstone Bridge

The Wheatstone bridge is an arrangement of four resistances used to determine one unknown resistance in terms of the other three. It consists of four arms $AB, BC, CD,$ and $DA$ containing resistances $P, Q, R,$ and $S$ respectively. A battery is connected across $A$ and $C$, and a sensitive galvanometer is connected between $B$ and $D$.

1.1 The Balanced Condition Derivation

A bridge is said to be "balanced" when no current flows through the galvanometer ($I_g = 0$).

Applying Kirchhoff’s Rules: Let $I$ be the total current. At junction $A$, it divides into $I_1$ (through $P$) and $I_2$ (through $R$). In the balanced state, the potential at point $B$ must equal the potential at point $D$ ($V_B = V_D$).

1. Potential Drop across $AB$ = Potential Drop across $AD$ $$V_A - V_B = V_A - V_D \implies I_1 P = I_2 R \quad \text{--- (Eq. 1)}$$
2. Since $I_g = 0$, the current through $BC$ is $I_1$ and through $CD$ is $I_2$ $$V_B - V_C = V_D - V_C \implies I_1 Q = I_2 S \quad \text{--- (Eq. 2)}$$
3. Dividing Eq. 1 by Eq. 2: $$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \implies \frac{P}{Q} = \frac{R}{S}$$

Key Inference: If three resistances are known, the fourth can be calculated. Usually, $P$ and $Q$ are "ratio arms," $R$ is a "known variable resistance," and $S$ is the "unknown resistance."

1.2 Sensitivity of Wheatstone Bridge

The bridge is most sensitive (shows maximum deflection for small changes in resistance) when all four resistances are of the same order of magnitude. If $P \approx Q \approx R \approx S$, the precision of the result is highest.

2. The Meter Bridge (Slide Wire Bridge)

The Meter Bridge is the practical application of the Wheatstone Bridge principle. It consists of a $1\text{ m}$ (100 cm) long manganin or constantan wire of uniform cross-section stretched on a wooden board.

2.1 Construction and Working

• Wire: Made of alloys with high resistivity and low temperature coefficient of resistance.
• Jockey: A metal slided used to make contact with the wire to find the null point.
• Circuit: The unknown resistance $X$ is connected in one gap, and a resistance box $R$ is connected in the other.

As per the Wheatstone principle, at the null point (distance $l$ from the left end $A$): $$\frac{R}{X} = \frac{\text{Resistance of length } l}{\text{Resistance of length } (100 - l)}$$ Since Resistance $\propto$ Length (for uniform wire): $$\frac{R}{X} = \frac{l}{100 - l} \implies X = R \left( \frac{100 - l}{l} \right)$$

2.2 End Corrections

In a physical meter bridge, the wire ends are soldered to copper strips. These strips and the contact points have their own resistances which are not accounted for in the $100\text{ cm}$ scale. These are called End Corrections ($\alpha$ and $\beta$).

• Let $\alpha$ be the equivalent length of the left end contact and $\beta$ for the right end.
• The corrected formula becomes: $$\frac{R}{X} = \frac{l + \alpha}{(100 - l) + \beta}$$

2.3 Minimizing Errors

3. The Potentiometer

The Potentiometer is an ideal voltmeter because it measures potential difference in a "null-point" condition, meaning it draws no current from the circuit being measured.

3.1 Potential Gradient ($k$)

The fundamental principle of a potentiometer is that the potential drop across a segment of a uniform wire is directly proportional to its length ($V \propto l$), provided the current is constant. $$V = I \sigma l \implies V = kl$$ Where $k = \frac{V_{wire}}{L_{total}}$ is the Potential Gradient (Unit: V/m). $$k = \left( \frac{E_p}{R_h + r_p + R_w} \right) \frac{R_w}{L}$$ (Here $E_p$ = primary cell EMF, $r_p$ = internal resistance of primary cell, $R_h$ = rheostat, $R_w$ = wire resistance).

3.2 Application: Comparison of EMFs ($E_1$ vs $E_2$)

Two cells of EMF $E_1$ and $E_2$ are connected via a two-way key.

1. When $E_1$ is in the circuit, null point is at $l_1$: $E_1 = kl_1$
2. When $E_2$ is in the circuit, null point is at $l_2$: $E_2 = kl_2$
3. Ratio: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$

3.3 Application: Measurement of Internal Resistance ($r$)

1. Open Circuit: Key in the resistance box circuit is open. Balance length $l_1$ gives $E = kl_1$.
2. Closed Circuit: A known resistance $S$ is connected across the cell. Balance length $l_2$ gives Terminal Voltage $V = kl_2$.
3. Formula: Since $r = \left( \frac{E}{V} - 1 \right) S$ $$r = S \left( \frac{l_1}{l_2} - 1 \right)$$

3.4 Maximizing Sensitivity

A potentiometer is "sensitive" if it can detect very small potential differences. Sensitivity can be increased by:

1. Decreasing Potential Gradient ($k$): This is done by increasing the length of the wire or decreasing the current in the primary circuit using a rheostat.
2. Sensitivity $\propto \frac{1}{k}$.

4. Comparison Summary Table

5. Concept Mnemonics & Tips

• Mnemonic for Potentiometer: "More Length, More Sensitive" ($k = V/L$, so as $L \uparrow$, $k \downarrow$, sensitivity $\uparrow$).
• The "Left" Rule: When calculating meter bridge problems, always ensure $l$ is measured from the end connected to the first resistance mentioned in your ratio.
• Standardization: The process of determining the potential gradient $k$ using a standard cell (like a Weston Cadmium cell) is called standardization.

6. Solved Numerical Examples

Example 1: Balanced Wheatstone Bridge

In a Wheatstone bridge, the four arms have resistances $P=10\Omega, Q=100\Omega, R=40\Omega$ and $S$ unknown. If the bridge is balanced, find $S$. If the battery is $12\text{V}$, what is the equivalent resistance?

Solution:

1. Balanced Condition: $\frac{P}{Q} = \frac{R}{S}$ $\frac{10}{100} = \frac{40}{S} \implies S = \frac{100 \times 40}{10} = 400\Omega$.
2. Equivalent Resistance ($R_{eq}$): In balance, $I_g=0$, so $(P+Q)$ is in parallel with $(R+S)$. $R_{top} = 10 + 100 = 110\Omega$ $R_{bottom} = 40 + 400 = 440\Omega$ $R_{eq} = \frac{110 \times 440}{110 + 440} = \frac{48400}{550} = 88\Omega$.

Example 2: Meter Bridge with End Correction

In a Meter Bridge, the null point is found at $40\text{ cm}$. When a $10\Omega$ resistor is added in series with the unknown resistance $X$ (in the right gap), the null point shifts to $25\text{ cm}$. Find $X$. (Assume $R$ in left gap remains constant).

Solution: Case 1: $\frac{R}{X} = \frac{40}{60} = \frac{2}{3} \implies R = \frac{2}{3}X$ Case 2: $\frac{R}{X+10} = \frac{25}{75} = \frac{1}{3}$ Substitute $R$: $\frac{2X/3}{X+10} = \frac{1}{3} \implies \frac{2X}{X+10} = 1 \implies 2X = X + 10 \implies X = 10\Omega$.

Example 3: Potentiometer Gradient Calculation

A potentiometer wire has a length of $10\text{ m}$ and resistance $20\Omega$. It is connected in series with a $2\text{V}$ battery and a $30\Omega$ external resistance. Calculate the potential gradient.

Solution:

1. Total Resistance of Primary Circuit: $R_{total} = R_{wire} + R_{ext} = 20 + 30 = 50\Omega$.
2. Current in Wire: $I = \frac{E}{R_{total}} = \frac{2}{50} = 0.04\text{ A}$.
3. Voltage across Wire: $V_w = I \times R_{wire} = 0.04 \times 20 = 0.8\text{ V}$.
4. Potential Gradient ($k$): $k = \frac{V_w}{L} = \frac{0.8}{10} = 0.08\text{ V/m}$ or $0.0008\text{ V/cm}$.

Example 4: Comparing Cells in Series

Two cells of EMF $E_1$ and $E_2$ ($E_1 > E_2$) are used in a potentiometer. When connected in "assist" mode ($E_1+E_2$), the balance length is $6\text{ m}$. When connected in "oppose" mode ($E_1-E_2$), the balance length is $2\text{ m}$. Find $E_1/E_2$.

Solution: We know $E \propto l$.

1. $E_1 + E_2 = k(6)$
2. $E_1 - E_2 = k(2)$ Dividing the equations: $\frac{E_1 + E_2}{E_1 - E_2} = \frac{6}{2} = 3$ $E_1 + E_2 = 3E_1 - 3E_2 \implies 4E_2 = 2E_1 \implies \frac{E_1}{E_2} = \frac{2}{1}$.

Example 5: Internal Resistance calculation

A cell gives a balance point at $75\text{ cm}$. When a $5\Omega$ resistor is connected across it, the balance point shifts to $60\text{ cm}$. Find the internal resistance $r$.

Solution: Given $l_1 = 75\text{ cm}$, $l_2 = 60\text{ cm}$, $S = 5\Omega$. $r = S \left( \frac{l_1}{l_2} - 1 \right)$ $r = 5 \left( \frac{75}{60} - 1 \right) = 5 \left( \frac{5}{4} - 1 \right) = 5 \left( \frac{1}{4} \right) = 1.25\Omega$.

Example 6: The "Short Wire" Error

In a potentiometer, the driver cell has EMF $2\text{V}$. A student tries to measure the EMF of a cell of $2.5\text{V}$. Explain why the student will not find a null point.

Solution: For a null point to exist, the potential drop across the potentiometer wire ($V_{wire}$) must be greater than or equal to the EMF to be measured ($E_{test}$). Here, $V_{wire} \leq E_{driver} \implies V_{wire} \leq 2\text{V}$. Since $E_{test} = 2.5\text{V}$ is greater than the maximum available potential drop on the wire ($2\text{V}$), the jockey will show deflection in the same direction at both ends of the wire. No point exists where $V_{wire_segment} = 2.5\text{V}$.

7. Errors and Precautions in Experiments

1. Jockey Sliding: Do not slide the jockey with pressure on the wire. This causes "kinking" or abrasion, which makes the cross-sectional area non-uniform, thereby making $k$ variable.
2. Current Limits: High currents cause Joule heating ($H = I^2Rt$), which increases the temperature of the wire and changes its resistance.
3. Driving Cell EMF: The EMF of the primary (driving) cell must always be greater than the EMF of the secondary cell(s), otherwise a null point cannot be found on the wire.
4. All-Positive Connection: Ensure the positive terminals of all cells (primary and secondary) are connected to the same zero-end of the potentiometer wire. If polarities are mixed, the potential will not "cancel out" to zero.

Heating Effects, Power & RC Circuits

The flow of electricity is not merely a transport of charge; it is a transport of energy. In the realm of JEE Main Physics, understanding how this energy is converted into heat and how capacitors modulate this flow over time is pivotal. This section explores the thermal manifestations of current and the transient behavior of circuits containing both resistors and capacitors.

1. Heating Effect of Electric Current (Joule’s Law)

When an electric current $I$ flows through a conductor of resistance $R$ for a time $t$, the electrical energy is converted into thermal energy (heat). This occurs because the drifting electrons undergo frequent collisions with the lattice ions of the conductor, transferring their kinetic energy to the lattice.

1.1 The Mathematical Formulation

According to Joule’s Law of Heating, the heat produced $H$ is:

1. Directly proportional to the square of the current ($H \propto I^2$) for a given $R$ and $t$.
2. Directly proportional to the resistance ($H \propto R$) for a given $I$ and $t$.
3. Directly proportional to the time ($H \propto t$) for which current flows.

Combining these, we get: $$H = I^2 R t$$

Using Ohm’s Law ($V = IR$), we can derive alternate forms: $$H = V I t = \frac{V^2}{R} t$$

Units:

• SI Unit: Joule (J)
• CGS Unit: Calorie (cal), where $1 \text{ cal} \approx 4.18 \text{ J}$.
• Heat in calories: $H = \frac{I^2 R t}{4.18}$

2. Electrical Power

Power is the rate at which work is done or energy is dissipated in an electrical circuit.

2.1 Theoretical Definition

$$P = \frac{dH}{dt} = \frac{dW}{dt}$$ From Joule’s Law: $$P = I^2 R = VI = \frac{V^2}{R}$$

2.2 Terminal Voltage vs. EMF Power

In a circuit with a battery of EMF $\varepsilon$ and internal resistance $r$:

• Power generated by the source: $P_{gen} = \varepsilon I$
• Power dissipated in internal resistance: $P_{int} = I^2 r$
• Power delivered to external load $R$: $P_{ext} = I^2 R = \frac{\varepsilon^2 R}{(R+r)^2}$

2.3 Commercial Unit of Energy

While the Joule is the SI unit, utility companies use the Kilowatt-hour (kWh), also known as a "Unit." $$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$$

Mnemonic: "PIV" and "VIP" Remember $P=IV$ as "Power is Very Important." In series, the bulb with the highest resistance glows brightest ($P \propto R$). In parallel, the bulb with the lowest resistance glows brightest ($P \propto 1/R$).

3. Maximum Power Transfer Theorem

For a network consisting of a source $(\varepsilon, r)$ and a variable external load $R$, the power delivered to the load is maximum when the external resistance is equal to the internal resistance of the source.

Derivation: Power $P = I^2 R = \left( \frac{\varepsilon}{R+r} \right)^2 R$. To maximize $P$, $\frac{dP}{dR} = 0$: $$\frac{d}{dR} \left[ \varepsilon^2 R (R+r)^{-2} \right] = 0$$ $$\varepsilon^2 \left[ (R+r)^{-2} - 2R(R+r)^{-3} \right] = 0$$ $$1 - \frac{2R}{R+r} = 0 \implies R+r = 2R \implies R = r$$

Maximum Power Value: $$P_{max} = \frac{\varepsilon^2 r}{(r+r)^2} = \frac{\varepsilon^2}{4r}$$ Efficiency at max power is only 50% because half the energy is lost in the internal resistance.

4. RC Circuits: The Transient State

In a purely resistive circuit, current reaches its steady state instantly. However, in circuits containing a capacitor ($C$), the charge and current vary with time as the capacitor stores energy.

4.1 Charging of a Capacitor

Consider a circuit with a DC source $V$, a resistor $R$, and an uncharged capacitor $C$ in series. At $t=0$, the switch is closed.

Derivation: Applying Kirchhoff’s Voltage Law (KVL): $$V - iR - \frac{q}{C} = 0$$ Since $i = \frac{dq}{dt}$: $$R \frac{dq}{dt} = V - \frac{q}{C} = \frac{CV - q}{C}$$ $$\int_0^q \frac{dq}{CV - q} = \int_0^t \frac{dt}{RC}$$ Let $q_0 = CV$ (Maximum charge). $$-\ln(q_0 - q) |_0^q = \frac{t}{RC}$$ $$\ln\left(\frac{q_0 - q}{q_0}\right) = -\frac{t}{RC}$$ $$q = q_0(1 - e^{-t/RC})$$

The Time Constant ($\tau$): The term $\tau = RC$ is the time constant.

• When $t = \tau$, $q = q_0(1 - e^{-1}) \approx 0.632 q_0$.
• Definition: Time taken to reach 63.2% of the maximum charge.

Current during charging: $$i = \frac{dq}{dt} = \frac{d}{dt} [q_0(1 - e^{-t/RC})] = \frac{q_0}{RC} e^{-t/RC}$$ $$i = I_0 e^{-t/RC}$$ (where $I_0 = V/R$)

4.2 Discharging of a Capacitor

If a capacitor with initial charge $Q_0$ is connected across a resistor $R$: Derivation: $$iR + \frac{q}{C} = 0 \implies R\frac{dq}{dt} = -\frac{q}{C}$$ $$\int_{Q_0}^q \frac{dq}{q} = -\int_0^t \frac{dt}{RC}$$ $$q = Q_0 e^{-t/RC}$$ Current: $i = \frac{dq}{dt} = -\frac{Q_0}{RC} e^{-t/RC}$ (The negative sign indicates current flow opposite to charging).

5. Energy Considerations in RC Charging

A curious phenomenon in electrostatics is that when charging a capacitor from a battery, exactly 50% of the energy supplied by the battery is dissipated as heat, regardless of the value of $R$.

Proof:

1. Work done by Battery ($W_b$): $W_b = \int dq \cdot V = V \int_0^{q_0} dq = V \cdot q_0 = V(CV) = CV^2$.
2. Energy Stored in Capacitor ($U_c$): $U_c = \frac{1}{2} C V^2$.
3. Heat Dissipated ($H$): $H = W_b - U_c = CV^2 - \frac{1}{2} CV^2 = \frac{1}{2} CV^2$.

Why is 50% lost? The resistance $R$ limits the current. The heat $H = \int i^2 R dt$. If $R \to 0$, the initial current $I_0 \to \infty$ for an infinitesimal time, and the integral still yields $\frac{1}{2} CV^2$. This energy is lost primarily as electromagnetic radiation or heat in the connecting wires.

Solved Problems

Problem 1: Power and Voltage Rating

A bulb is rated $100\text{ W}, 220\text{ V}$. If the line voltage drops to $110\text{ V}$, calculate the new power consumed. Solution: First, find the resistance of the bulb (assumed constant): $R = \frac{V_R^2}{P_R} = \frac{220^2}{100} = 484\text{ }\Omega$. New Power $P' = \frac{V'^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\text{ W}$. Quick Tip: If voltage becomes $1/n$, power becomes $1/n^2$. Here $n=2$, so $100/2^2 = 25$.

Problem 2: Combination of Bulbs

Two bulbs $P_1 = 60\text{ W}$ and $P_2 = 100\text{ W}$ (both rated for $V$) are connected in series across supply $V$. Which one is brighter? Solution: Resistance $R \propto 1/P_{rated}$. $R_1 = V^2/60$, $R_2 = V^2/100$. Clearly $R_1 > R_2$. In series, $P_{actual} = I^2 R$. Since $I$ is same, $P \propto R$. Bulb 1 ($60\text{ W}$) has higher resistance, so it will consume more power and glow brighter.

Problem 3: Maximum Power Transfer

A cell of EMF $6\text{ V}$ and internal resistance $2\text{ }\Omega$ is connected to a variable resistor $R$. Find the maximum power delivered and the current at this state. Solution: For $P_{max}$, $R = r = 2\text{ }\Omega$. Total resistance $R_{eq} = 2 + 2 = 4\text{ }\Omega$. Current $I = \varepsilon / R_{eq} = 6/4 = 1.5\text{ A}$. $P_{max} = I^2 R = (1.5)^2 \times 2 = 2.25 \times 2 = 4.5\text{ W}$. Using formula: $P_{max} = \varepsilon^2 / 4r = 36 / (4 \times 2) = 4.5\text{ W}$.

Problem 4: Time Constant of RC Circuit

A $2\text{ }\mu\text{F}$ capacitor is charged through a $1\text{ M}\Omega$ resistor. How much time is required for the charge to reach 86.5% of its final value? Solution: Given $q = 0.865 q_0$. Equation: $q = q_0(1 - e^{-t/\tau})$. $0.865 q_0 = q_0(1 - e^{-t/\tau}) \implies 0.865 = 1 - e^{-t/\tau} \implies e^{-t/\tau} = 0.135$. We know $e^{-1} \approx 0.37$, $e^{-2} \approx 0.135$. So, $t/\tau = 2 \implies t = 2 \tau$. $\tau = RC = 10^6 \times 2 \times 10^{-6} = 2\text{ s}$. $t = 2 \times 2 = 4\text{ s}$.

Problem 5: Internal Resistance and Heating

A battery of $10\text{ V}$ is short-circuited. If the heat produced in the battery is $20\text{ W}$, find the internal resistance. Solution: In short circuit, external $R = 0$. $I = \varepsilon / r$. Power dissipated in internal resistance $P = I^2 r = (\varepsilon/r)^2 r = \varepsilon^2 / r$. $20 = 10^2 / r \implies 20 = 100 / r \implies r = 5\text{ }\Omega$.

Problem 6: Energy in RC Charging

A $5\text{ }\mu\text{F}$ capacitor is connected to a $12\text{ V}$ battery via a $10\text{ k}\Omega$ resistor. Find the total heat dissipated during the complete charging process. Solution: As derived in Section 5, heat dissipated $H = \frac{1}{2}CV^2$. $H = 0.5 \times (5 \times 10^{-6}) \times (12)^2$ $H = 0.5 \times 5 \times 10^{-6} \times 144$ $H = 360 \times 10^{-6}\text{ J} = 0.36\text{ mJ}$. Note: The resistance value is irrelevant for the total heat.

Problem 7: Current in RC Circuit

In a charging RC circuit, $R=200\text{ }\Omega$, $C=100\text{ }\mu\text{F}$, and $V=2\text{ V}$. Find the current at $t=20\text{ ms}$. Solution: $\tau = RC = 200 \times 100 \times 10^{-6} = 0.02\text{ s} = 20\text{ ms}$. Instantaneous current $i = I_0 e^{-t/\tau}$. $I_0 = V/R = 2/200 = 0.01\text{ A}$. At $t = 20\text{ ms}$, $t = \tau$, so: $i = 0.01 \times e^{-1} = 0.01 \times 0.37 = 0.0037\text{ A} = 3.7\text{ mA}$.

Problem 8: Discharging Capacitor

A capacitor charged to $100\text{ V}$ is discharged through a resistor. After $1\text{ s}$, the voltage across it is $37\text{ V}$. Find the time constant. Solution: $V_c = V_0 e^{-t/\tau}$. $37 = 100 e^{-1/\tau} \implies 0.37 = e^{-1/\tau}$. Since $e^{-1} \approx 0.368 \approx 0.37$, we have: $1/\tau = 1 \implies \tau = 1\text{ s}$.

50 PYQ-Style JEE Main Problems

This section provides a rigorous collection of 50 problems modeled after the National Testing Agency (NTA) patterns for JEE Main. These problems integrate concepts from Electrostatics, Capacitance, and Current Electricity, focusing on the high-yield areas that frequently appear in the exam.

Problem Solving Strategies for Exam Day

Part 1: Electrostatics & Flux (Problems 1–15)

1. Point Charge and Flux: A point charge $+q$ is placed at the center of a cube of side $L$. What is the electric flux through one face of the cube?

• Solution: By Gauss’s Law, total flux $\phi_{total} = \frac{q}{\epsilon_0}$. Since a cube has 6 identical faces and the charge is at the center, the flux is distributed equally. Thus, $\phi_{face} = \frac{q}{6\epsilon_0}$.

2. Electric Field of a Ring: A ring of radius $R$ has a total charge $Q$ distributed uniformly. At what distance $x$ from the center on its axis is the electric field maximum?

• Solution: The field on the axis is $E = \frac{kQx}{(R^2+x^2)^{3/2}}$. To find the maximum, set $\frac{dE}{dx} = 0$. Using the quotient rule, we find $x = \frac{R}{\sqrt{2}}$.

3. Dipole in Non-uniform Field: An electric dipole is placed in a non-uniform electric field. What does it experience?

• Solution: In a non-uniform field, the forces on the two charges $+q$ and $-q$ do not cancel out. Therefore, it experiences both a net force and a net torque.

4. Surface Charge Density: Two spheres of radii $R_1$ and $R_2$ are connected by a wire and charged. What is the ratio of their surface charge densities $\sigma_1/\sigma_2$?

• Solution: When connected, their potentials become equal ($V_1 = V_2$). Thus, $\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \implies \frac{Q_1}{R_1} = \frac{Q_2}{R_2}$. Since $\sigma = \frac{Q}{4\pi R^2}$, then $\frac{\sigma_1}{\sigma_2} = \frac{Q_1/R_1^2}{Q_2/R_2^2} = \frac{R_2}{R_1}$.

5. Force Between Plates: A parallel plate capacitor has area $A$ and charge $Q$. What is the force of attraction between the plates?

• Solution: The field produced by one plate is $E = \frac{\sigma}{2\epsilon_0} = \frac{Q}{2A\epsilon_0}$. The force on the other plate is $F = QE = \frac{Q^2}{2A\epsilon_0}$.

6. Potential at Center: Four charges $+q, +q, -q, -q$ are placed at the corners of a square of side $a$. Find the potential at the center.

• Solution: Potential is a scalar. $V = \sum \frac{kq}{r}$. Since distances to the center are all equal ($r = a/\sqrt{2}$), and the sum of charges $(q + q - q - q) = 0$, the potential $V = 0$.

7. Work Done by Field: A charge $q$ is moved from point A to B in an electric field. If $V_A = 10V$ and $V_B = 25V$, find the work done by the external agent.

• Solution: $W_{ext} = q(V_B - V_A)$. If $q = 2\mu C$, then $W = 2 \times 10^{-6} \times (25-10) = 30 \times 10^{-6} = 30 \mu J$.

8. Infinite Sheet Flux: A square frame of side $10 cm$ is placed parallel to an infinite sheet of charge density $\sigma = 17.7 \times 10^{-12} C/m^2$. Find the flux through it.

• Solution: $E = \frac{\sigma}{2\epsilon_0}$. $\phi = E \cdot A = \frac{17.7 \times 10^{-12}}{2 \times 8.85 \times 10^{-12}} \times (0.1)^2 = 1 \times 0.01 = 0.01 Vm$.

9. Potential Energy of System: Three charges $Q, q, q$ are placed at equilibrium on a straight line at distances $0, r/2, r$. Find $Q$.

• Solution: For the charge $q$ at the end to be in equilibrium, $F_{net} = 0$. $\frac{kQq}{(r/2)^2} + \frac{kq^2}{r^2} = 0 \implies \frac{4kQq}{r^2} = -\frac{kq^2}{r^2} \implies Q = -q/4$.

10. Gauss’s Law Variable Field: The electric field in a region is $E = ax \hat{i}$. Find the charge enclosed in a cube of side $L$ with one corner at origin.

• Solution: $\phi_{in} = 0$ (at $x=0$). $\phi_{out} = E_{x=L} \cdot L^2 = (aL)L^2 = aL^3$. By Gauss Law, $q_{encl} = \epsilon_0 \phi = \epsilon_0 a L^3$.

11. Zero Potential Point: Two charges $+9e$ and $+e$ are $16 cm$ apart. Where is the electric field zero?

• Solution: Let $x$ be distance from $9e$. $\frac{k(9e)}{x^2} = \frac{ke}{(16-x)^2}$. Taking square root: $\frac{3}{x} = \frac{1}{16-x} \implies 48 - 3x = x \implies x = 12 cm$.

12. Charged Pendulum: A mass $m$ with charge $q$ hangs by a string in a horizontal field $E$. Find the tension $T$.

• Solution: At equilibrium, $T \cos \theta = mg$ and $T \sin \theta = qE$. Squaring and adding: $T = \sqrt{(mg)^2 + (qE)^2}$.

13. Shell Potential: A hollow conducting sphere of radius $R$ is given a charge $Q$. Plot $V$ vs $r$.

• Solution: Inside ($r < R$), $V = \frac{kQ}{R}$ (constant). Outside ($r > R$), $V = \frac{kQ}{r}$. The graph is a horizontal line followed by a hyperbola.

14. Kinetic Energy of Electron: An electron is accelerated through $100V$. Find its final velocity.

• Solution: $\frac{1}{2}mv^2 = eV$. $v = \sqrt{2eV/m} = \sqrt{2 \times 1.6 \times 10^{-19} \times 100 / 9.1 \times 10^{-31}} \approx 5.9 \times 10^6 m/s$.

15. Energy Stored in Field: If the electric field intensity is $E$, what is the energy density?

• Solution: $U_d = \frac{1}{2}\epsilon_0 E^2$. Unit is $J/m^3$.

Part 2: Capacitance (Problems 16–25)

16. Series/Parallel Mix: Three capacitors $2\mu F, 3\mu F, 6\mu F$ are in series. Find $C_{eq}$.

• Solution: $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$. So $C_{eq} = 1\mu F$.

17. Dielectric Slab Induction: A capacitor $C$ is charged to $V$ and then disconnected. A slab of $K=4$ is inserted. Find new Energy.

• Solution: Charge $Q$ is constant. $U_{old} = \frac{1}{2}CV^2$. $U_{new} = \frac{Q^2}{2C_{new}} = \frac{Q^2}{2KC} = \frac{U_{old}}{K}$. Energy decreases by factor of 4.

18. Battery Connected Dielectric: A capacitor remains connected to a $10V$ battery while a dielectric $K=2$ is inserted. What happens to charge?

• Solution: $V$ is constant. $Q_{new} = (KC)V = K Q_{old}$. Charge doubles.

19. Common Potential: A $10\mu F$ capacitor charged to $50V$ is connected to an uncharged $20\mu F$ capacitor. Find common potential.

• Solution: $V_{common} = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} = \frac{10 \times 50 + 20 \times 0}{10 + 20} = \frac{500}{30} = 16.67 V$.

20. Energy Loss in Sharing: Calculate energy loss in Problem 19.

• Solution: $\Delta U = \frac{C_1C_2(V_1 - V_2)^2}{2(C_1 + C_2)} = \frac{10 \times 20 (50-0)^2}{2(30)} \times 10^{-6} J = \frac{200 \times 2500}{60} \times 10^{-6} \approx 8.33 mJ$.

21. Force on Dielectric: A slab is partially inserted into a capacitor. Explain the force.

• Solution: The fringing fields at the edges of the plates polarize the dielectric, creating a force that pulls the slab into the capacitor to minimize potential energy.

22. Spherical Capacitor: Find capacitance of Earth ($R = 6400 km$).

• Solution: $C = 4\pi\epsilon_0 R = \frac{6.4 \times 10^6}{9 \times 10^9} \approx 711 \mu F$.

23. Parallel Plates with Multiple Slabs: A capacitor has two slabs of thickness $d/2$ with $K_1$ and $K_2$.

• Solution: They act as two capacitors in series. $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$ where $C_1 = \frac{K_1 \epsilon_0 A}{d/2}$ and $C_2 = \frac{K_2 \epsilon_0 A}{d/2}$.

24. Work Done to Increase Plate Separation: A capacitor $C$ with charge $Q$ has separation $d$. Work to move plates to $2d$?

• Solution: $W = U_f - U_i = \frac{Q^2}{2C_f} - \frac{Q^2}{2C_i}$. Since $C_f = C/2$, $W = \frac{Q^2}{2(C/2)} - \frac{Q^2}{2C} = \frac{Q^2}{2C}$.

25. N-Identical Drops: $n$ drops each of radius $r$ and potential $v$ merge into one big drop. Find new potential $V$.

• Solution: Charge conserved: $Q = nq$. Volume conserved: $R = n^{1/3}r$. $V = \frac{kQ}{R} = \frac{k(nq)}{n^{1/3}r} = n^{2/3} v$.

Part 3: Current Electricity & Circuits (Problems 26–40)

26. Drift Velocity: If current $I$ is doubled and area $A$ is halved, what happens to $v_d$?

• Solution: $I = nAev_d \implies v_d = \frac{I}{nAe}$. $v_{d}' = \frac{2I}{n(A/2)e} = 4v_d$. It quadruples.

27. Temperature Coefficient: Resistance of a wire is $5\Omega$ at $50^\circ C$ and $6\Omega$ at $100^\circ C$. Find $R$ at $0^\circ C$.

• Solution: $R = R_0(1+\alpha \Delta T)$.
  1. $5 = R_0(1 + 50\alpha)$
  2. $6 = R_0(1 + 100\alpha)$ Divide (2) by (1): $\frac{6}{5} = \frac{1+100\alpha}{1+50\alpha} \implies 6 + 300\alpha = 5 + 500\alpha \implies 1 = 200\alpha \implies \alpha = 0.005/^\circ C$. Substitute back: $5 = R_0(1 + 50 \times 0.005) \implies 5 = R_0(1.25) \implies R_0 = 4\Omega$.

28. Kirchhoff’s Loop Rule: A $10V$ battery is connected to two resistors $2\Omega$ and $3\Omega$ in series. Find potential drop across $3\Omega$.

• Solution: $I = \frac{V}{R_{eq}} = \frac{10}{2+3} = 2A$. $V_{3\Omega} = IR = 2 \times 3 = 6V$.

29. Wheatstone Bridge: In a balanced bridge, $P=10, Q=20, R=5$. Find $S$.

• Solution: $P/Q = R/S \implies 10/20 = 5/S \implies S = 10\Omega$.

30. Potentiometer Sensitivity: How can you increase the sensitivity of a potentiometer?

• Solution: By decreasing the potential gradient ($k = V/L$). This is done by increasing the length of the wire $L$ or decreasing the current in the primary circuit using a rheostat.

31. Color Coding: A resistor has bands Yellow, Violet, Brown, Gold. Find value.

• Solution: Yellow(4), Violet(7), Brown($10^1$), Gold($\pm 5%$). Value = $470 \Omega \pm 5%$.

32. Internal Resistance: A cell of emf $1.5V$ gives $0.2A$ through a $7\Omega$ resistor. Find $r$.

• Solution: $I = \frac{E}{R+r} \implies 0.2 = \frac{1.5}{7+r} \implies 1.4 + 0.2r = 1.5 \implies 0.2r = 0.1 \implies r = 0.5\Omega$.

33. Power Rating: Two bulbs $60W$ and $100W$ are in series. Which glows brighter?

• Solution: In series, $I$ is same. $P = I^2 R$. Higher resistance glows brighter. $R = V^2/P_{rated}$. $60W$ bulb has higher resistance ($R_{60} > R_{100}$), so it glows brighter.

34. Meter Bridge: Balance point is at $40 cm$ from left. If $X$ is in left gap and $R=60\Omega$ in right, find $X$.

• Solution: $\frac{X}{40} = \frac{60}{100-40} \implies \frac{X}{40} = \frac{60}{60} = 1 \implies X = 40\Omega$.

35. Combination of Cells: $n$ cells of emf $E$ and internal resistance $r$ are in series. If one cell is reversed, find new $E$ and $r$.

• Solution: Net $E = (n-1)E - E = (n-2)E$. Net $r = nr$ (resistance is independent of polarity).

36. Current in Mesh: (Conceptual) In a circuit with multiple batteries, how do you handle signs?

• Solution: Use $\sum \Delta V = 0$. If moving from $-$ to $+$ terminal of battery, $\Delta V = +E$. If moving in direction of current through resistor, $\Delta V = -IR$.

37. Potentiometer Balance: A cell is balanced at $250 cm$. When shunted by $5\Omega$, balance point becomes $200 cm$. Find $r$.

• Solution: $r = R\left(\frac{l_1}{l_2} - 1\right) = 5\left(\frac{250}{200} - 1\right) = 5(1.25 - 1) = 1.25\Omega$.

38. Maximum Power Theorem: A battery of $E, r$ is connected to external $R$. Power in $R$ is max when?

• Solution: When $R = r$. Max power $P_{max} = \frac{E^2}{4r}$.

39. Stretching a Wire: A wire of $10\Omega$ is stretched to triple its length. New $R$?

• Solution: $R \propto L^2$ (since volume $AL$ is constant). $R' = (3)^2 R = 9 \times 10 = 90\Omega$.

40. Galvanometer to Ammeter: $G=100\Omega, I_g = 1mA$. How to measure $1A$?

• Solution: Parallel shunt $S = \frac{I_g G}{I - I_g} = \frac{0.001 \times 100}{1 - 0.001} \approx \frac{0.1}{1} = 0.1\Omega$.

Part 4: RC Circuits & Advanced Problems (Problems 41–50)

41. Charging RC: A $2\mu F$ capacitor is charged through $1M\Omega$ by a $10V$ battery. Find charge at $t=2s$.

• Solution: $\tau = RC = 10^6 \times 2 \times 10^{-6} = 2s$. $q = q_0(1 - e^{-t/\tau})$. At $t=2s$, $q = q_0(1 - e^{-1}) = 20\mu C(1 - 0.37) = 12.6\mu C$.

42. Discharging RC: A capacitor $C$ has potential $V$. It is discharged through $R$. Find time to reach $V/2$.

• Solution: $V = V_0 e^{-t/RC} \implies \frac{1}{2} = e^{-t/RC} \implies \ln 2 = t/RC \implies t = RC \ln 2$.

43. Current at $t=0$: In an RC circuit, what is the current the instant the switch is closed?

• Solution: The capacitor behaves like a short circuit (plain wire). $I = V/R$.

44. Steady State Capacitor: In a complex DC circuit, a branch has a capacitor. How much current flows there in steady state?

• Solution: Zero. The capacitor acts as an open circuit ($R_c = \infty$).

45. Energy Lost in RC Charging: How much energy is dissipated as heat in the resistor while charging a capacitor to energy $U$?

• Solution: Total energy supplied by battery = $QV = CV^2$. Energy stored in capacitor = $\frac{1}{2}CV^2$. Energy lost as heat = $CV^2 - \frac{1}{2}CV^2 = \frac{1}{2}CV^2$. Ratio is 1:1.

46. Two-Capacitor Circuit: $C_1$ charged to $V_1$ is connected via $R$ to $C_2$ (uncharged). Find current at $t=0$.

• Solution: $I_0 = \frac{V_1}{R}$. The second capacitor $C_2$ acts as a wire initially.

47. Equivalent Time Constant: Two capacitors $C_1, C_2$ in parallel connected to $R$ in series.

• Solution: $\tau = R(C_1 + C_2)$.

48. Nodal Analysis: Three wires meet at a junction with potentials $10V, 20V, 30V$ through resistors $1\Omega, 2\Omega, 3\Omega$. Find junction potential $V_0$.

• Solution: $\frac{V_0 - 10}{1} + \frac{V_0 - 20}{2} + \frac{V_0 - 30}{3} = 0$. Multiply by 6: $6V_0 - 60 + 3V_0 - 60 + 2V_0 - 60 = 0 \implies 11V_0 = 180 \implies V_0 = 16.36V$.

49. Symmetrical Cube Resistance: 12 resistors of $R$ form edges of a cube. Find $R_{eq}$ across body diagonal.

• Solution: $R_{eq} = \frac{5}{6}R$. (Mnemonic: 3-6-2 edges approach for current).

50. Power in AC/DC Comparison: A resistor $R$ produces heat $H$ with DC current $I$. What peak AC current $I_0$ produces same heat?

• Solution: $I_{rms} = I$. Since $I_{rms} = I_0/\sqrt{2}$, then $I_0 = I\sqrt{2}$.

Key Formulas Summary Table

Formula Sheet & Last-15-Days Plan

This section is designed to be your final battlefield manual. We have condensed the vast expanse of Electrostatics, Capacitance, and Current Electricity into high-yield formulas, a rigorous schedule, and technical nuances that differentiate a 95 percentile score from a 99.9 percentile score.

H3: The Ultimate Formula Compendium (80+ High-Yield Formulas)

In JEE Main, speed is as crucial as accuracy. Categorized below are the essential mathematical tools for Electrostatics and Current Electricity.

1. Electrostatics: Point Charges & Fields

1. Coulomb’s Law: $F = k \frac{q_1 q_2}{r^2}$ where $k = \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 , \text{Nm}^2\text{C}^{-2}$.
2. Force in Medium: $F_m = \frac{F_{vac}}{K}$ (where $K$ is Dielectric Constant).
3. Electric Field ($E$): $\vec{E} = \frac{\vec{F}}{q_0}$.
4. Field of Point Charge: $E = \frac{kq}{r^2}$.
5. Electric Dipole Moment: $\vec{p} = q \cdot (2\vec{l})$ (Directed $-q$ to $+q$).
6. Field on Axial Line of Dipole: $E_{axial} = \frac{2kp r}{(r^2 - l^2)^2} \approx \frac{2kp}{r^3}$ (for $r \gg l$).
7. Field on Equatorial Line: $E_{eq} = \frac{kp}{(r^2 + l^2)^{3/2}} \approx \frac{kp}{r^3}$ (for $r \gg l$).
8. General Point Field ($r, \theta$): $E = \frac{kp}{r^3} \sqrt{1 + 3\cos^2\theta}$.
9. Torque on Dipole: $\vec{\tau} = \vec{p} \times \vec{E} = pE\sin\theta$.
10. Potential Energy of Dipole: $U = -\vec{p} \cdot \vec{E} = -pE\cos\theta$.
11. Flux ($\Phi$): $\Phi = \int \vec{E} \cdot d\vec{A} = \frac{q_{encl}}{\epsilon_0}$ (Gauss’s Law).

2. Continuous Charge Distributions

12. Infinite Line Charge: $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
13. Infinite Sheet of Charge: $E = \frac{\sigma}{2\epsilon_0}$.
14. Conducting Plate Field: $E = \frac{\sigma}{\epsilon_0}$.
15. Uniformly Charged Ring: $E_x = \frac{kqx}{(x^2 + R^2)^{3/2}}$ (Max at $x = \frac{R}{\sqrt{2}}$).
16. Spherical Shell ($r > R$): $E = \frac{kq}{r^2}$.
17. Spherical Shell ($r < R$): $E = 0$.
18. Solid Non-conducting Sphere ($r > R$): $E = \frac{kq}{r^2}$.
19. Solid Non-conducting Sphere ($r < R$): $E = \frac{kqr}{R^3}$.

3. Electric Potential & Energy

20. Potential ($V$): $V = \frac{W_{\infty \to P}}{q}$.
21. Potential of Point Charge: $V = \frac{kq}{r}$.
22. Relation $E$ and $V$: $E = -\frac{dV}{dr}$ or $\vec{E} = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k})$.
23. Potential of Dipole: $V = \frac{kp\cos\theta}{r^2}$.
24. Potential Energy (System of 2 charges): $U = \frac{kq_1q_2}{r}$.
25. Self Energy (Shell): $U_{self} = \frac{kq^2}{2R}$.
26. Self Energy (Solid Sphere): $U_{self} = \frac{3kq^2}{5R}$.
27. Equipotential surfaces: $\vec{E} \perp$ Surface.

4. Capacitance and Dielectrics

28. Definition: $C = \frac{Q}{V}$.
29. Parallel Plate Capacitor (PPC): $C = \frac{\epsilon_0 A}{d}$.
30. PPC with Dielectric: $C = \frac{K\epsilon_0 A}{d}$.
31. Spherical Capacitor: $C = 4\pi\epsilon_0 \frac{ab}{b-a}$.
32. Isolated Sphere Capacitance: $C = 4\pi\epsilon_0 R$.
33. Cylindrical Capacitor: $C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$.
34. Series Combination: $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}$.
35. Parallel Combination: $C_{eq} = \sum C_i$.
36. Energy Stored: $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$.
37. Energy Density: $u_e = \frac{1}{2}\epsilon_0 E^2$.
38. Force between plates: $F = \frac{Q^2}{2\epsilon_0 A}$.
39. Dielectric Insertion (Battery Disconnected): $Q$ stays constant, $V \to V/K$, $E \to E/K$, $C \to KC$, $U \to U/K$.
40. Dielectric Insertion (Battery Connected): $V$ stays constant, $Q \to KQ$, $E$ stays same, $C \to KC$, $U \to KU$.

5. Current Electricity: Fundamentals

41. Current: $I = \frac{dq}{dt} = neAv_d$.
42. Drift Velocity: $v_d = \frac{eE\tau}{m} = \frac{eV\tau}{ml}$.
43. Current Density: $J = \frac{I}{A} = \sigma E$.
44. Mobility: $\mu = \frac{v_d}{E}$.
45. Ohm’s Law: $V = IR$ or $\vec{J} = \sigma \vec{E}$.
46. Resistance: $R = \rho \frac{l}{A}$.
47. Conductivity ($\sigma$): $\sigma = \frac{1}{\rho}$.
48. Temperature Dependence: $R_t = R_0(1 + \alpha \Delta T)$.
49. Resistivity in terms of $\tau$: $\rho = \frac{m}{ne^2\tau}$.

6. DC Circuits & Kirchoof’s Laws

50. Series Resistors: $R_{eq} = R_1 + R_2 + \dots$
51. Parallel Resistors: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$
52. KCL (Junction Rule): $\sum I_{in} = \sum I_{out}$.
53. KVL (Loop Rule): $\sum \Delta V = 0$.
54. EMF and Internal Resistance: $V = \mathcal{E} - Ir$ (Discharging), $V = \mathcal{E} + Ir$ (Charging).
55. Cells in Series: $\mathcal{E}_{eq} = \sum \mathcal{E}i$, $r{eq} = \sum r_i$.
56. Cells in Parallel (Identical): $\mathcal{E}{eq} = \mathcal{E}$, $r{eq} = r/n$.
57. Cells in Parallel (Different): $\mathcal{E}_{eq} = \frac{\frac{\mathcal{E}_1}{r_1} + \frac{\mathcal{E}_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$.
58. Max Power Transfer Theorem: Power to load is max when $R_{ext} = r_{int}$.
59. Wheatstone Bridge Balance: $\frac{P}{Q} = \frac{R}{S}$.

7. Measuring Instruments

60. Galvanometer to Ammeter: $S = \frac{I_g G}{I - I_g}$ (Shunt in parallel).
61. Galvanometer to Voltmeter: $R = \frac{V}{I_g} - G$ (Resistance in series).
62. Potentiometer Gradient: $k = \frac{V}{L} = \left(\frac{\mathcal{E}}{R+r+R_s}\right) \frac{\rho}{A}$.
63. Potentiometer (Compare EMF): $\frac{\mathcal{E}_1}{\mathcal{E}_2} = \frac{l_1}{l_2}$.
64. Potentiometer (Internal Resistance): $r = R \left(\frac{l_1 - l_2}{l_2}\right)$.
65. Meter Bridge: $X = R \frac{l}{100 - l}$.

8. RC Circuits (Transient State)

66. Charging Charge: $q(t) = Q_0(1 - e^{-t/RC})$.
67. Charging Current: $i(t) = \frac{V}{R}e^{-t/RC}$.
68. Discharging Charge: $q(t) = Q_0 e^{-t/RC}$.
69. Time Constant: $\tau = RC$ (Time to reach 63.2% charge).

9. Electrical Power

70. Power: $P = VI = I^2 R = \frac{V^2}{R}$.
71. Heat Produced (Joule’s Law): $H = I^2 R t$.
72. Bulb Brightness: Brightness $\propto$ Actual Power.
73. Series Bulbs: $P_{total} = (\frac{1}{P_1} + \frac{1}{P_2})^{-1}$ (Lower wattage bulb glows brighter).
74. Parallel Bulbs: $P_{total} = P_1 + P_2$ (Higher wattage bulb glows brighter).

10. Advanced Electrostatics (Specific Cases)

75. Work done to move charge: $W = q(V_{final} - V_{initial})$.
76. Force on Dielectric Slabs: $F = \frac{\epsilon_0 b (K-1) V^2}{2d}$.
77. Electric Field in cavity of conductor: $E = 0$.
78. Pressure on Conductive surface: $P_{elec} = \frac{\sigma^2}{2\epsilon_0}$.
79. Velocity of charge through potential difference: $v = \sqrt{\frac{2qV}{m}}$.
80. Radius of curvature of trajectory in $E$ field: $R = \frac{mv^2}{qE\sin\theta}$.

H3: 15-Day Revision Strategy (The "Polishing" Phase)

This plan assumes you have finished the syllabus once and are now in the final lap before JEE Main.

H3: Worked Examples (Application of Theory)

Example 1: The Nodal Method in DC Circuits

Problem: Find the current through the $2\Omega$ resistor in a bridge where a $10V$ source is connected across a complex network with no clear series/parallel.

Solution:

1. Identify Nodes: Label one node as $0V$ (Ground).
2. Assign Potentials: Let the unknown node potential be $x$.
3. KCL Equation: $\sum \frac{V_x - V_i}{R_i} = 0$.
4. Calculations: Say $\frac{x-10}{R_1} + \frac{x-0}{R_2} + \frac{x-V_y}{R_3} = 0$.
5. Solve for x: Find Current $I = \frac{\Delta V}{R}$. Tip: Nodal analysis reduces the number of equations compared to Loop Law.

Example 2: Energy in Variable Dielectrics

Problem: A PPC of area $A$ and distance $d$ is filled with a dielectric $K(x) = K_0(1 + \frac{x}{d})$. Find its capacitance.

Step-by-Step:

1. Identify Symmetry: The dielectric varies with $x$ (thickness), so consider a small element $dx$ as a capacitor.
2. Series/Parallel? These slabs are in series.
3. Incremental Capacitance: $dC = \frac{K(x)\epsilon_0 A}{dx}$.
4. Integration: $\frac{1}{C_{eq}} = \int \frac{1}{dC} = \int_{0}^{d} \frac{dx}{K_0(1 + \frac{x}{d})\epsilon_0 A}$.
5. Final Result: $\frac{1}{C_{eq}} = \frac{d}{K_0 \epsilon_0 A} \ln(2) \Rightarrow C_{eq} = \frac{K_0 \epsilon_0 A}{d \ln 2}$.

H3: Common Pitfalls: Integration & Graphs

1. Integration Limits: When calculating potential from field ($V = -\int \vec{E} \cdot d\vec{r}$), always check if the reference point is $\infty$ or the surface.
2. The "R" vs "r" Trap: In solid spheres, the field increases linearly inside ($E \propto r$) and decreases quadratically outside ($E \propto 1/r^2$). Students often swap these in exams.
3. Short-Circuiting Capacitors: If a wire is connected across a capacitor in steady state, the capacitor is bypassed. However, in the transient state ($t=0$), it acts like a wire itself.
4. Internal Resistance in Instruments: Many forget to add the internal resistance of the battery when calculating the potential gradient $(k)$ of a potentiometer. Remember: $k = \frac{I \rho}{A}$ where $I$ depends on the entire circuit.

H3: 40-Point Final Checklist for Success

1. Can you define Electric Flux without using the word "lines"?
2. Do you know that $E$ is zero inside a conductor only in electrostatic equilibrium?
3. Can you find the neutral point between two unlike charges? (Outside, near the smaller magnitude).
4. Have you memorized the $E$-field of a disk at its axis?
5. Do you remember that $\vec{E}$ is always perpendicular to equipotential surfaces?
6. Is the difference between "Potential" and "Potential Energy" clear?
7. Can you calculate the work done to rotate a dipole from $0^\circ$ to $180^\circ$? ($2pE$).
8. Do you know that Gauss Law is valid for all closed surfaces, but useful only for symmetric ones?
9. Can you prove why a charge stays on the outer surface of a conductor?
10. Do you know the graph of $V$ vs $r$ for a hollow sphere? (Constant inside).
11. Have you practiced the "Method of Images" for basic problems?
12. Can you calculate the capacitance of a PPC with multiple vertical dielectric slabs? (Parallel).
13. Do you understand that $Q = CV$, where $V$ is potential difference, not potential?
14. Can you find the equivalent capacitance of an infinite ladder?
15. Do you know why a dielectric increases capacitance? (Polarization reduces $E_{net}$).
16. Have you solved problems where a dielectric is partially pulled out?
17. Do you know the force on a dielectric slab is $F = \frac{\partial U}{\partial x}$?
18. Is the concept of "Redistribution of Charges" clear? (Potential becomes equal).
19. Do you know the energy lost during charge sharing? $\Delta U = \frac{C_1 C_2 (V_1-V_2)^2}{2(C_1+C_2)}$.
20. Can you differentiate between $I = n e A v_d$ and $J = \sigma E$?
21. Do you know why Ohm’s law is not a universal law? (Diodes, Semiconductors).
22. Have you mastered the color code table? (BB ROY of Great Britain...).
23. Can you calculate $R_{eq}$ for a cube across its body diagonal? ($5R/6$).
24. Across face diagonal? ($3R/4$).
25. Across one edge? ($7R/12$).
26. Do you know the difference between EMF and Terminal Voltage?
27. Can you solve a Wheatstone bridge when it is unbalanced? (Nodal analysis).
28. Is the condition for maximum power transfer clear ($R=r$)?
29. Do you understand why a Voltmeter must have high resistance?
30. ...and why an Ammeter must have low resistance?
31. Can you explain the "null point" concept in a Meter Bridge?
32. Do you know that the Potentiometer is an "ideal" voltmeter because it draws no current at balance?
33. Can you convert a $0-1mA$ galvanometer to a $0-10A$ ammeter?
34. Do you know the behavior of an inductor vs a capacitor at $t=0$ and $t=\infty$?
35. Is the formula for the time constant $\tau = RC$ etched in your brain?
36. Can you sketch the curve for $q$ vs $t$ for a discharging capacitor?
37. Do you know that $P = V^2/R$ is best for parallel and $P=I^2R$ for series?
38. Have you reviewed the "Heating effect" in a wire with variable cross-section?
39. Is the concept of "Shorting" and "Open Circuit" visually intuitive for you?
40. Have you practiced at least 15 years of JEE Main PYQs for these two units?

Pro-Tip: In the last 15 days, do not read new theory. Every hour spent solving a "newly discovered" complex HC Verma problem is an hour lost in revising NCERT-based facts that JEE Main actually tests. Stick to the sheet. Trust the process.