PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 1 · JEE Main

Electric Charges and Fields

Chapter-wise practice and revision for JEE Main: concise notes on charge and Coulomb's law, electric fields of point and continuous charges, dipoles, and flux and Gauss's law, concept figures, and 51 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

51 selected questions4 topics7 concept figuresPreview · first 3 of 51 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 143 previous-year JEE questions in the PiTheory bank

Gauss's law54
Electric field35
Coulomb's law27
Dipole24

Contents

  1. 1Electric charge and Coulomb's lawQ1–Q12
  2. 2Electric field of point and continuous chargesQ13–Q25
  3. 3Electric dipoleQ26–Q36
  4. 4Electric flux and Gauss's lawQ37–Q51
  5. ✓Last-day checklist
01

Topic 1 · Q1–Q12

Electric charge and Coulomb's law

27 of the 143 JEE PYQs in the bank test this as their main idea

Look first

++rFFlike charges repelF = kq1q2 / (K r²)k = 9×10⁹ N m² C⁻²forces add as vectors
Coulomb's law: inverse square, along the line joining the charges.

Quick notes

$$\vec F_{12}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}\,\hat r_{12},\qquad k=\frac{1}{4\pi\varepsilon_0}=9\times10^9\ \text{N m}^2\text{C}^{-2},\qquad F_{\text{medium}}=\frac{F_{\text{vacuum}}}{K}$$

Charge is quantised ($q=ne$, $e=1.6\times10^{-19}$ C), conserved, and additive. Forces from several charges add as vectors (superposition).

SituationResult
Identical conductors touchedcharges average: $\dfrac{q_1+q_2}{2}$ each
$Q$ split as $q$, $Q-q$ for maximum force$q=Q/2$
Zero force on a third charge, like charges $q_1,q_2$between them, $\dfrac{x}{d-x}=\sqrt{q_1/q_2}$
Same, unlike chargesoutside, beyond the smaller charge
Two hanging balls, small angle$x^3=\dfrac{q^2l}{2\pi\varepsilon_0mg}$
Same angle in a liquid$K=\dfrac{\rho}{\rho-\rho_l}$
Drops merging ($n$ drops)$R=n^{1/3}r$, $\sigma\to n^{1/3}\sigma$
Charge between two equal charges, small displacementalong line: $\omega^2=\dfrac{4kQq}{ma^3}$; perpendicular (opposite signs): $\omega^2=\dfrac{2kQq}{ma^3}$
  • Same force in a medium as in air at distance $d$: equivalent air distance $d\sqrt K$.
  • Electric force between an electron and a proton is about $10^{39}$ times the gravitational force.
  • Non-uniform charge density: $Q=\int\rho\,dV$, with $dV=4\pi r^2dr$ for spheres.
Traps. Units ($\mu$C, cm). Coulomb's law is for point charges; for spheres use the centre-to-centre distance only when they are far apart.

Practice: first 3 of 12 questions

Q1JEE Main 2022 (26 Jun, Shift 2)Easy

Sixty four conducting drops each of radius $0.02\ \text{m}$ and each carrying a charge of $5\ \mu\text{C}$ are combined to form a bigger drop. The ratio of surface density of the bigger drop to the smaller drop will be

  1. A$1 : 4$
  2. B$4 : 1$
  3. C$1 : 8$
  4. D$8 : 1$
Answer and solution

Answer: Option B: $4 : 1$

Concept. When drops merge, charge adds and volume adds. Surface density is $\sigma=\dfrac{Q}{4\pi R^2}$.

Step 1. New radius. Volume $\times64$, so radius $\times4$: $R=4r$.

Step 2. New charge. $Q=64q$.

Step 3. Ratio. $\dfrac{\sigma_{\text{big}}}{\sigma_{\text{small}}}=\dfrac{64q/(4r)^2}{q/r^2}=\dfrac{64}{16}=4$.

Answer. $4:1$ (Option B).

Point to note. $\sigma\propto n^{1/3}$: $64^{1/3}=4$.

Q2JEE Main 2019 (9 Jan, Shift 2)Medium

Charge is distributed within a sphere of radius $R$ with a volume charge density $\rho(r) = \dfrac{A}{r^2}e^{-2r/a}$, where $A$ and $a$ are constants. If $Q$ is the total charge of this charge distribution, the radius $R$ is:

  1. A$\dfrac a2\log\left(\dfrac{1}{1-\dfrac{Q}{2\pi aA}}\right)$
  2. B$a\log\left(\dfrac{1}{1-\dfrac{Q}{2\pi aA}}\right)$
  3. C$a\log\left(1-\dfrac{Q}{2\pi aA}\right)$
  4. D$\dfrac a2\log\left(1-\dfrac{Q}{2\pi aA}\right)$
Answer and solution

Answer: Option A: $\dfrac a2\log\left(\dfrac{1}{1-\dfrac{Q}{2\pi aA}}\right)$

Concept. Total charge of a non-uniform sphere: add thin shells, $dQ=\rho(r)\,4\pi r^2dr$.

Step 1. Set up. $Q=\displaystyle\int_0^R\frac{A}{r^2}e^{-2r/a}\,4\pi r^2\,dr=4\pi A\int_0^Re^{-2r/a}dr$.

Step 2. Integrate. $Q=4\pi A\cdot\dfrac a2\left(1-e^{-2R/a}\right)=2\pi aA\left(1-e^{-2R/a}\right)$.

Step 3. Solve for $R$. $e^{-2R/a}=1-\dfrac{Q}{2\pi aA}\Rightarrow R=\dfrac a2\log\left(\dfrac{1}{1-\frac{Q}{2\pi aA}}\right)$.

Answer. Option A.

Point to note. $dQ=\rho\,4\pi r^2dr$.

Q3JEE Main 2025 (24 Jan, Shift 2)Easy

A small uncharged conducting sphere is placed in contact with an identical sphere but having $4\times10^{-8}\ \text{C}$ charge and then removed to a distance such that the force of repulsion between them is $9\times10^{-3}\ \text{N}$. The distance between them is (Take $\dfrac{1}{4\pi\epsilon_0}$ as $9\times10^9$ in SI units)

  1. A$3\ \text{cm}$
  2. B$2\ \text{cm}$
  3. C$4\ \text{cm}$
  4. D$1\ \text{cm}$
Answer and solution

Answer: Option B: $2\ \text{cm}$

Concept. Identical conductors in contact share charge equally. Then Coulomb's law $F=\dfrac{kq_1q_2}{r^2}$ gives the distance.

Step 1. Charges after contact. $2\times10^{-8}$ C each.

Step 2. Distance. $9\times10^{-3}=\dfrac{9\times10^9\times(2\times10^{-8})^2}{r^2}=\dfrac{3.6\times10^{-6}}{r^2}\Rightarrow r^2=4\times10^{-4}$, $r=0.02$ m.

Answer. $2$ cm (Option B).

Point to note. Contact halves the charge.

48 more questions and 3 more topics in the full chapter

  1. 2Electric field of point and continuous charges13 questions with solutions
  2. 3Electric dipole11 questions with solutions
  3. 4Electric flux and Gauss's law15 questions with solutions
  4. ✓Last-day checklist

The complete Electric Charges and Fields 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.

Best value

Complete Pack

Class 11 + Class 12: all 28 chapters in one organised package.

₹799₹449
  • 28 chapters, all included
  • 1,346 selected JEE Main questions (up to 50 per chapter)
  • Concept notes, equations and methods
  • Step-by-step solutions, answer keys, difficulty levels, checklists
  • Save ₹149 on buying both class packs

Class 12 Pack

All 14 Class 12 chapters: electrostatics to semiconductors.

₹399₹299
  • 14 chapters, all included
  • 695 selected JEE Main questions (up to 50 per chapter)
  • Concept notes, equations and methods
  • Step-by-step solutions, answer keys, difficulty levels, checklists
  1. Tap Buy on WhatsAppYour message opens with the pack you chose. Send it.
  2. Get the payment detailsSreeraj Sir replies with the UPI details on WhatsApp.
  3. Pay and share the screenshotAny UPI app works: GPay, PhonePe, Paytm.
  4. Receive your PDFYour personal copy, with your name on it, comes on WhatsApp.