Electrostatics & Potential Theory: Master Notes

Chapter Overview

Electrostatics forms the foundation of electromagnetism. In IIT-JEE (Advanced), top problems focus on Gauss's Law with symmetrical and non-symmetrical cavities, electrostatic energy densities ($u_E=\frac{1}{2}{\epsilon }_0{E}^2$), conductor charge redistribution, and dipole interaction potentials.

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1. Coulomb's Law, Electric Fields & Potentials

$$\overrightarrow{F}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}\hat{r},\overrightarrow{E}=-\mathrm{\nabla }V=-(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k})$$$$V(\overrightarrow{r})=-{\int }_{\mathrm{\infty }}^{\overrightarrow{r}}\overrightarrow{E}\cdot d\overrightarrow{r}$$

1.1 High-Yield Standard Field & Potential Configurations

Charge Distribution	Electric Field $\overrightarrow{E}$	Electric Potential $V$
Infinitely Long Wire ($\lambda$)	$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$	$V=-\frac{\lambda }{2\pi {\epsilon }_0}\ln \left(\frac{r}{r_0}\right)$
Infinite Flat Sheet ($\sigma$)	$E=\frac{\sigma }{2{\epsilon }_0}$ (Independent of distance)	$V=-\frac{\sigma }{2{\epsilon }_0}x+V_0$
Uniform Ring of Radius $R$ on Axis	$E(x)=\frac{1}{4\pi {\epsilon }_0}\frac{Qx}{(R^2+x^2{)}^{3/2}}$	$V(x)=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{\sqrt{R^2+x^2}}$
Uniform Circular Disk on Axis	$E(x)=\frac{\sigma }{2{\epsilon }_0}(1-\frac{x}{\sqrt{R^2+x^2}})$	$V(x)=\frac{\sigma }{2{\epsilon }_0}(\sqrt{R^2+x^2}-x)$
Uniform Solid Sphere (Radius $R$, Total $Q$)	$E_{\text{in}}=\frac{Qr}{4\pi {\epsilon }_0{R}^3},E_{\text{out}}=\frac{Q}{4\pi {\epsilon }_0{r}^2}$	$V_{\text{in}}=\frac{Q}{8\pi {\epsilon }_0{R}^3}(3R^2-r^2),V_0=1.5V_s$

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2. Gauss's Law & Cavity Problems

$${\mathrm{\Phi }}_E={\oint }_{\mathcal{S}}\overrightarrow{E}\cdot d\overrightarrow{A}=\frac{Q_{\text{enclosed}}}{{\epsilon }_0}$$

2.1 The Cavity Theorem (Uniform Electric Field inside a Spherical Cavity)

For a non-conducting sphere with uniform charge density $\rho$ having a spherical cavity whose center is displaced by vector $\overrightarrow{a}$ from the sphere's center:

Multi-Mode DiagramRotational Dynamics: Spool on Incline & Pure Rolling

Option 1: Publication-Grade Scientific Vector SVG

Free-body force decomposition on incline showing normal force N, static friction fs, tension T, and Instantaneous Axis of Rotation C.

Linear Acceleration: $a_{\text{cm}} = \frac{g\sin\theta}{1 + I_{\text{cm}}/(MR^2)} = \frac{2}{3}g\sin\theta$

Critical Tension Angle: $\phi_c = \arccos(r/R)$

Option 2: Interactive 60 FPS Canvas Physics Engine

Real-time 60 FPS pure rolling simulation on incline.

Incline Angle θ: 25 °

Acceleration ($a_{\text{cm}}$):2.76 m/s²

Option 3: Interactive 3D Spatial Viewport

3D rotatable gyroscope disc.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Precession Equation:$\vec{\tau} = \frac{d\vec{L}}{dt}$

Option 5: Reactive Parametric Physics with Live Calculus

Parametric angle sliders with live torque calculation.

Incline Angle θ: 25 °

Pull Angle φ: 45 °

Critical Angle ($\phi_c$):$62.5°$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

$${\overrightarrow{E}}_{\text{cavity}}={\overrightarrow{E}}_{\text{solid}}-{\overrightarrow{E}}_{\text{removed}}=\frac{\rho {\overrightarrow{r}}_1}{3{\epsilon }_0}-\frac{\rho {\overrightarrow{r}}_2}{3{\epsilon }_0}=\frac{\rho ({\overrightarrow{r}}_1-{\overrightarrow{r}}_2)}{3{\epsilon }_0}$$$$\boxed{{\displaystyle {\overrightarrow{E}}_{\text{cavity}}=\frac{\rho \overrightarrow{a}}{3{\epsilon }_0}=\text{constant in magnitude and direction everywhere inside!}}}$$

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3. Electrostatic Energy & Conductor Pressure

• Energy Density in Electric Field:$$u_E=\frac{1}{2}{\epsilon }_0{E}^2$$
• Electrostatic Pressure on Conductor Surface:$$P_e=\frac{1}{2}{\epsilon }_0{E}_{\text{surface}}^2=\frac{{\sigma }^2}{2{\epsilon }_0}$$
• Self-Energy of Spherical Shell ($Q,R$): $U=\frac{Q^2}{8\pi {\epsilon }_{0}R}$
• Self-Energy of Uniform Solid Sphere ($Q,R$): $U=\frac{3Q^2}{20\pi {\epsilon }_{0}R}$

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 (Paper 2) — Cavity Field and Potential Difference

Question:
A solid non-conducting sphere of radius $R$ has a uniform charge density $\rho$. A spherical cavity of radius $R/2$ is carved out, with its center located at distance $R/2$ from the sphere's center. Find:

1. The electric field vector inside the cavity.
2. The potential difference $V(O_1)-V(O_2)$ between the sphere center $O_1$ and the cavity center $O_2$.

Step-by-Step Solution:

• 1. Electric Field Inside the Cavity: Using the cavity theorem:

  $$\overrightarrow{E}=\frac{\rho \overrightarrow{a}}{3{\epsilon }_0}=\frac{\rho (R/2\hat{i})}{3{\epsilon }_0}=\boxed{{\displaystyle \frac{\rho R}{6{\epsilon }_0}\hat{i}}}$$

  (The field is completely uniform!)

• 2. Potential Difference: Since $\overrightarrow{E}$ is constant along the line connecting $O_1$ and $O_2$:

  $$V(O_1)-V(O_2)={\int }_{O_1}^{O_2}\overrightarrow{E}\cdot d\overrightarrow{r}=E\cdot d=\left(\frac{\rho R}{6{\epsilon }_0}\right)\left(\frac{R}{2}\right)=\boxed{{\displaystyle \frac{\rho R^2}{12{\epsilon }_0}}}$$

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PYQ 2: JEE Advanced 2020 — Concentric Conducting Shells

Question:
Three concentric conducting spherical shells $A,B,C$ have radii $R,2R,3R$ respectively. Shells $A$ and $C$ are given charges $+q$ and $-2q$, while shell $B$ is earthed. Find the charge $q_B$ induced on shell $B$.

Step-by-Step Solution:

1. Shell $B$ is connected to earth $⟹V_B=0$.
2. The potential at radius $r=2R$ is the sum of potentials due to charges on all three shells:$$V_B=\frac{1}{4\pi {\epsilon }_0}(\frac{q_A}{2R}+\frac{q_B}{2R}+\frac{q_C}{3R})=0$$
3. Substitute $q_A=+q$ and $q_C=-2q$:$$\frac{q}{2R}+\frac{q_B}{2R}-\frac{2q}{3R}=0$$$$\frac{q_B}{2R}=\frac{2q}{3R}-\frac{q}{2R}=\frac{4q-3q}{6R}=\frac{q}{6R}⟹\boxed{{\displaystyle q_B=\frac{q}{3}}}$$

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5. High-Yield Formula Sheet

Entity	Formula	Notes
Dipole Potential	$V=\frac{1}{4\pi {\epsilon }_0}\frac{p\cos \theta }{r^2}$	Far field $r\gg d$
Dipole Torque & Energy	$\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E},U=-\overrightarrow{p}\cdot \overrightarrow{E}$	Stable at $\theta =0$
Conductor Pressure	$P=\frac{{\sigma }^2}{2{\epsilon }_0}$	Always outward
Solid Sphere Self-Energy	$U=\frac{3Q^2}{20\pi {\epsilon }_{0}R}$	Work to assemble charge
Cavity Field	$\overrightarrow{E}=\frac{\rho \overrightarrow{a}}{3{\epsilon }_0}$	Uniform inside cavity