PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 2 · JEE Main

Electrostatic Potential and Capacitance

Chapter-wise practice and revision for JEE Main: concise notes on potential and potential energy, conductors, capacitance and combinations, dielectrics, and stored energy, concept figures, and 53 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

53 selected questions6 topics5 concept figuresPreview · first 3 of 53 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 181 previous-year JEE questions in the PiTheory bank

Dielectrics40
Potential27
Energy26
Capacitance20
Conductors19
Potential energy16

Contents

  1. 1Electric potential and its relation to the fieldQ1–Q9
  2. 2Potential energy and workQ10–Q16
  3. 3Conductors, shells and equipotential surfacesQ17–Q24
  4. 4Capacitance and combinationsQ25–Q32
  5. 5DielectricsQ33–Q42
  6. 6Energy stored, sharing of charge and chargingQ43–Q53
  7. ✓Last-day checklist
01

Topic 1 · Q1–Q9

Electric potential and its relation to the field

27 of the 181 JEE PYQs in the bank test this as their main idea

Look first

V = kq/requipotentials ⊥ field lines
Equipotentials are spheres; field lines cross them at right angles.

Quick notes

$$V=\frac{kq}{r},\qquad \vec E=-\nabla V\ \left(E_x=-\frac{\partial V}{\partial x}\right),\qquad V_B-V_A=-\int_A^B\vec E\cdot d\vec l$$

Potential is a scalar: add the potentials of all charges with their signs. In a uniform field, $\Delta V=-E\,d$ along the field.

SourcePotential
Ring or any arc, at its centre$\dfrac{kQ}{R}$ (shape does not matter)
Ring, on axis at $x$$\dfrac{kQ}{\sqrt{R^2+x^2}}$
Shell$\dfrac{kQ}{R}$ inside and on; $\dfrac{kQ}{r}$ outside
Solid sphere (uniform)$\dfrac{kQ(3R^2-r^2)}{2R^3}$ inside; centre $\tfrac32V_{\text{surface}}$
Dipole$\dfrac{kp\cos\theta}{r^2}$; zero on the equator
Line charge (difference)$V_1-V_2=2k\lambda\ln\dfrac{r_2}{r_1}$
  • Merging $n$ drops: $Q\times n$, $R\times n^{1/3}$, $V\times n^{2/3}$, energy $\times n^{5/3}$, $\sigma\times n^{1/3}$.
  • Given $V(r)$: $E=-dV/dr$, then Gauss's law gives the enclosed charge.
  • For positive charges only, $V$ can never be zero at a finite point, but $E$ can.
Traps. $V$ adds as scalars, $E$ as vectors. Sign in $\vec E=-\nabla V$.

Practice: first 3 of 9 questions

Q1JEE Main 2026 (6 Apr, Shift 2)Easy

The electric potential as a function of $x, y$ is given by $V = 5(x^2 - y^2)$ V. The electric field at a point $(2, 3)$ m is ______ V/m.

  1. A$(-20\hat{i} + 30\hat{j})$
  2. B$(20\hat{i} - 30\hat{j})$
  3. C$(20\hat{i} + 45\hat{j})$
  4. D$(-4\hat{i} + 6\hat{j})$
Answer and solution

Answer: Option A: $(-20\hat{i} + 30\hat{j})$

Concept. The field is the negative gradient of the potential: $E_x=-\dfrac{\partial V}{\partial x}$, $E_y=-\dfrac{\partial V}{\partial y}$.

Step 1. Components. $E_x=-10x$, $E_y=+10y$.

Step 2. At $(2,3)$. $\vec E=-20\hat i+30\hat j$ V/m.

Answer. Option A.

Point to note. $\vec E=-\nabla V$.

Q2JEE Main 2019 (8 Apr, Shift 2)Easy

The electric field in a region is given by $\vec E = (Ax + B)\hat i$, where $E$ is in $\text{N C}^{-1}$ and $x$ is in metres. The values of constants are $A = 20$ SI unit and $B = 10$ SI unit. If the potential at $x = 1$ is $V_1$ and that at $x = -5$ is $V_2$, then $V_1 - V_2$ is

  1. A$320\ \text{V}$
  2. B$-520\ \text{V}$
  3. C$180\ \text{V}$
  4. D$-48\ \text{V}$
Answer and solution

Answer: Option C: $180\ \text{V}$

Concept. $V_1-V_2=-\displaystyle\int_{x_2}^{x_1}E\,dx$ (potential falls along the field).

Step 1. Integrate. $\displaystyle\int_{-5}^{1}(20x+10)\,dx=\left[10x^2+10x\right]_{-5}^{1}=20-200=-180$.

Step 2. Sign. $V_1-V_2=-(-180)=180$ V.

Answer. Option C.

Point to note. $V_1-V_2=-\int E\,dx$.

Q3JEE Main 2017 (8 Apr)Medium

There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $P$, in the region, is found to vary between the limits $589.0$ V to $589.8$ V. What is the potential at a point on the sphere whose radius vector makes an angle of $60^\circ$ with the direction of the field?

  1. A$589.4$ V
  2. B$589.5$ V
  3. C$589.2$ V
  4. D$589.6$ V
Answer and solution

Answer: Option C: $589.2$ V

Concept. In a uniform field, $V=V_P-Er\cos\theta$ on a small sphere of radius $r$ about P. The maximum and minimum are $V_P\pm Er$.

Step 1. Centre value and spread. $V_P=\dfrac{589.0+589.8}{2}=589.4$ V, $Er=0.4$ V.

Step 2. At $60^\circ$. $589.4-0.4\times\tfrac12=589.2$ V.

Answer. Option C.

Point to note. $V=V_P-Er\cos\theta$.

50 more questions and 5 more topics in the full chapter

  1. 2Potential energy and work7 questions with solutions
  2. 3Conductors, shells and equipotential surfaces8 questions with solutions
  3. 4Capacitance and combinations8 questions with solutions
  4. 5Dielectrics10 questions with solutions
  5. 6Energy stored, sharing of charge and charging11 questions with solutions
  6. ✓Last-day checklist

The complete Electrostatic Potential and Capacitance chapter is in the Class 12 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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