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📘 Electrostatics — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Electrostatics for JEE Main with previous year questions on Coulomb's law, the electric field, and Gauss's law. Each solution is explained in a simple, step-by-step manner to boost your score.
Electrostatics — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (8 April Shift 2)
Two metal spheres of radius \(R\) and \(3R\) have same surface charge density \(\sigma\). If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes \(\sigma_1\) and \(\sigma_2\), respectively. The ratio \(\frac{\sigma_1}{\sigma_2}\) is :
(1) \(\frac{1}{9}\) (2) 9
(3) \(\frac{1}{3}\) (4) 3
(1) \(\frac{1}{9}\) (2) 9
(3) \(\frac{1}{3}\) (4) 3
✅ Answer & Explanation:
Correct Option: (4)
- When spheres are brought in contact, they reach the same potential \(V\).
- \(V = \frac{kQ_1}{R} = \frac{kQ_2}{3R} \implies \frac{Q_1}{R} = \frac{Q_2}{3R} \implies Q_2 = 3Q_1\).
- Surface charge densities:
- \(\sigma_1 = \frac{Q_1}{4\pi R^2}\)
- \(\sigma_2 = \frac{Q_2}{4\pi (3R)^2} = \frac{3Q_1}{36\pi R^2} = \frac{Q_1}{12\pi R^2}\).
- Ratio \(\frac{\sigma_1}{\sigma_2} = \frac{Q_1 / 4\pi R^2}{Q_1 / 12\pi R^2} = \frac{12}{4} = 3\).
Q2. JEE Main 2025 (8 April Shift 2)
An electron is released from rest near an infinite non-conducting sheet of uniform charge density \(-\sigma\). The rate of change of de-Broglie wave length associated with the electron varies inversely as \(n^{th}\) power of time. The numerical value of \(n\) is ________.
✅ Answer & Explanation:
Correct Answer: 2
- Field due to sheet is constant: \(E = \frac{\sigma}{2\epsilon_0}\).
- Force on electron \(F = eE = \text{constant}\). Acceleration \(a = \frac{e\sigma}{2m\epsilon_0} = \text{constant}\).
- Velocity at time \(t\) starting from rest: \(v = at\).
- de-Broglie wavelength \(\lambda = \frac{h}{mv} = \frac{h}{mat} \implies \lambda \propto t^{-1}\).
- Rate of change \(\frac{d\lambda}{dt} \propto \frac{d(t^{-1})}{dt} \propto -t^{-2}\).
- This varies inversely as the 2nd power of time, so \(n = 2\).
Q3. JEE Main 2025 (24 Jan Shift 2)
A small uncharged conducting sphere is placed in contact with an identical sphere but having \(4 \times 10^{-8} \, \text{C}\) charge and then removed to a distance such that the force of repulsion between them is \(9 \times 10^{-3} \, \text{N}\). The distance between them is :
(1) 3 cm (2) 2 cm
(3) 4 cm (4) 1 cm
(1) 3 cm (2) 2 cm
(3) 4 cm (4) 1 cm
✅ Answer & Explanation:
Correct Option: (2)
- Upon contact, charge shares equally: \(q_1 = q_2 = 2 \times 10^{-8} \, \text{C}\).
- \(F = \frac{kq^2}{r^2} \implies 9 \times 10^{-3} = \frac{9 \times 10^9 \times (4 \times 10^{-16})}{r^2}\).
- \(r^2 = \frac{36 \times 10^{-7}}{9 \times 10^{-3}} = 4 \times 10^{-4} \implies r = 2 \times 10^{-2} \, \text{m} = 2 \, \text{cm}\).
Q4. JEE Main 2025 (22 Jan Shift 1)
An electron enters symmetrically between metal plates of 10 cm length with horizontal velocity \(10^6 \, \text{m/s}\). If \(E = 9.1 \, \text{V/cm}\), the vertical component of velocity of electron emerging out is :
(1) 0 (2) \(1 \times 10^6 \, \text{m/s}\)
(3) \(16 \times 10^6 \, \text{m/s}\) (4) \(16 \times 10^4 \, \text{m/s}\)
(1) 0 (2) \(1 \times 10^6 \, \text{m/s}\)
(3) \(16 \times 10^6 \, \text{m/s}\) (4) \(16 \times 10^4 \, \text{m/s}\)
✅ Answer & Explanation:
Correct Option: (3)
- Time to cross plates \(t = L/v_x = 0.1 / 10^6 = 10^{-7} \, \text{s}\).
- Vertical acceleration \(a_y = eE/m = (1.6 \times 10^{-19} \times 910) / 9.1 \times 10^{-31} = 1.6 \times 10^{14} \, \text{m/s}^2\).
- \(v_y = a_y t = 1.6 \times 10^{14} \times 10^{-7} = 1.6 \times 10^7 = 16 \times 10^6 \, \text{m/s}\).
Q5. JEE Main 2022 (29 Jun Shift 2)
Electric potential \(V = 3x^2 \, \text{V}\). The electric field at point (1, 0, 3) m will be :
(1) \(3 \, \text{V/m}\), along +x (2) \(3 \, \text{V/m}\), along -x
(3) \(6 \, \text{V/m}\), along -x (4) \(6 \, \text{V/m}\), along +x
(1) \(3 \, \text{V/m}\), along +x (2) \(3 \, \text{V/m}\), along -x
(3) \(6 \, \text{V/m}\), along -x (4) \(6 \, \text{V/m}\), along +x
✅ Answer & Explanation:
Correct Option: (3)
- \(\vec{E} = -\frac{dV}{dx} \hat{i} = -\frac{d(3x^2)}{dx} \hat{i} = -6x \hat{i}\).
- At \(x = 1\), \(\vec{E} = -6 \hat{i} \, \text{V/m}\).
- Magnitude is 6 V/m, directed along the negative x-axis.
Q6. JEE Main 2021 (25 Feb Shift 2)
Two identical spheres have \(2.1 \, \text{nC}\) and \(-0.1 \, \text{nC}\) charges. They are brought in contact and separated by 0.5 m. The force between them is _____ \(\times 10^{-9} \, \text{N}\).
✅ Answer & Explanation:
Correct Answer: 36
- Shared charge \(q = (2.1 - 0.1)/2 = 1.0 \, \text{nC}\).
- \(F = \frac{kq^2}{r^2} = \frac{9 \times 10^9 \times (10^{-9})^2}{(0.5)^2} = \frac{9 \times 10^{-9}}{0.25} = 36 \times 10^{-9} \, \text{N}\).
Q7. JEE Main 2025 (7 April Shift 2)
A dipole with two electric charges of \(2 \, \mu\text{C}\) magnitude each, with separation distance \(0.5 \, \mu\text{m}\), is placed between the plates of a capacitor such that its axis is parallel to an electric field establishing between the plates when a potential difference of 5 V is applied. Separation between the plates is 0.5 mm. If the dipole is rotated by \(30^\circ\) from the axis, the value of torque is :
(1) \(5 \times 10^{-9} \, \text{Nm}\) (2) \(5 \times 10^{-3} \, \text{Nm}\)
(3) \(2.5 \times 10^{-12} \, \text{Nm}\) (4) \(2.5 \times 10^{-9} \, \text{Nm}\)
(1) \(5 \times 10^{-9} \, \text{Nm}\) (2) \(5 \times 10^{-3} \, \text{Nm}\)
(3) \(2.5 \times 10^{-12} \, \text{Nm}\) (4) \(2.5 \times 10^{-9} \, \text{Nm}\)
✅ Answer & Explanation:
Correct Option: (1)
- Dipole moment \(p = qd = (2 \times 10^{-6}) \times (0.5 \times 10^{-6}) = 10^{-12} \, \text{C}\cdot\text{m}\).
- Electric field \(E = \frac{V}{r} = \frac{5}{0.5 \times 10^{-3}} = 10^4 \, \text{V/m}\).
- Torque \(\tau = pE \sin\theta = (10^{-12}) \times (10^4) \times \sin 30^\circ\).
- \(\tau = 10^{-8} \times 0.5 = 5 \times 10^{-9} \, \text{Nm}\).
Q8. JEE Main 2025 (29 Jan Shift 1)
An electric dipole of mass m, charge q, and length \(l\) is placed in a uniform electric field \(\vec{E} = E_0 \hat{i}\). When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be:
(1) \(\frac{1}{2\pi} \sqrt{\frac{ml}{2qE_0}}\) (2) \(2\pi \sqrt{\frac{ml}{qE_0}}\)
(3) \(\frac{1}{2\pi} \sqrt{\frac{2ml}{qE_0}}\) (4) \(2\pi \sqrt{\frac{ml}{2qE_0}}\)
(1) \(\frac{1}{2\pi} \sqrt{\frac{ml}{2qE_0}}\) (2) \(2\pi \sqrt{\frac{ml}{qE_0}}\)
(3) \(\frac{1}{2\pi} \sqrt{\frac{2ml}{qE_0}}\) (4) \(2\pi \sqrt{\frac{ml}{2qE_0}}\)
✅ Answer & Explanation:
Correct Option: (4)
- Restoring torque \(\tau = -pE \sin\theta \approx -pE \theta\).
- \(I\alpha = -pE\theta \implies \alpha = -(pE/I)\theta\).
- For two point masses \(\frac{m}{2}\) separated by \(l\), \(I = 2 \times (\frac{m}{2})(\frac{l}{2})^2 = ml^2/4\).
- Dipole moment \(p = ql\).
- \(\omega^2 = \frac{ql E_0}{ml^2 / 4} = \frac{4qE_0}{ml} \implies T = 2\pi\sqrt{\frac{ml}{4qE_0}} = 2\pi\sqrt{\frac{ml}{2qE_0}}\).
Q9. JEE Main 2025 (28 Jan Shift 2)
An electric dipole of dipole moment \(6 \times 10^{-6} \, \text{Cm}\) is placed in uniform electric field of magnitude \(10^6 \, \text{V/m}\). Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ____ J.
✅ Answer & Explanation:
Correct Answer: 12
- Work done \(W = \Delta U = U_f - U_i\).
- \(U_i = -pE \cos 0^\circ = -pE\).
- \(U_f = -pE \cos 180^\circ = +pE\).
- \(W = pE - (-pE) = 2pE = 2 \times (6 \times 10^{-6}) \times (10^6) = 12 \, \text{J}\).
Q10. JEE Main 2025 (23 Jan Shift 2)
Two charges \(7 \, \mu\text{c}\) and \(-4 \, \mu\text{c}\) are placed at (-7 cm, 0, 0) and (7 cm, 0, 0) respectively. The electrostatic potential energy of the configuration is :
(1) -1.8 J (2) -2.0 J
(3) -1.5 J (4) -1.2 J
(1) -1.8 J (2) -2.0 J
(3) -1.5 J (4) -1.2 J
✅ Answer & Explanation:
Correct Option: (1)
- Distance \(r = 7 - (-7) = 14 \, \text{cm} = 0.14 \, \text{m}\).
- \(U = \frac{kq_1 q_2}{r} = \frac{9 \times 10^9 \times (7 \times 10^{-6}) \times (-4 \times 10^{-6})}{0.14}\).
- \(U = \frac{-252 \times 10^{-3}}{0.14} = -1.8 \, \text{J}\).
Q11. JEE Main 2025 (22 Jan Shift 2)
For a short dipole moment \(P\) along x-axis at origin, potential and field at A \((r, 0)\) are \(V_0\) and \(E_0\). The correct combination of potential and field at point B \((0, 2r)\) on y-axis is :
(1) \(V_0\) and \(E_0/4\) (2) zero and \(E_0/16\)
(3) zero and \(E_0/8\) (4) \(V_0/2\) and \(E_0/16\)
(1) \(V_0\) and \(E_0/4\) (2) zero and \(E_0/16\)
(3) zero and \(E_0/8\) (4) \(V_0/2\) and \(E_0/16\)
✅ Answer & Explanation:
Correct Option: (2)
- Point A is axial: \(V_0 = \frac{kp}{r^2}\), \(E_0 = \frac{2kp}{r^3}\).
- Point B is equatorial: Potential \(V_B = 0\) (standard result).
- Field at B \((2r)\): \(E_B = \frac{kp}{(2r)^3} = \frac{kp}{8r^3}\).
- Since \(kp/r^3 = E_0/2\), then \(E_B = (E_0/2)/8 = E_0/16\).
Q12. JEE Main 2023 (06 Apr Shift 2)
A dipole has mass \(m\) for positive charge and \(2m\) for negative charge, separated by distance \(l\). In field \(E\), the angular frequency of small oscillations is :
(1) \(\sqrt{\frac{3qE}{2ml}}\) (2) \(\sqrt{\frac{8qE}{ml}}\)
(3) \(\sqrt{\frac{4qE}{ml}}\) (4) \(\sqrt{\frac{8qE}{3ml}}\)
(1) \(\sqrt{\frac{3qE}{2ml}}\) (2) \(\sqrt{\frac{8qE}{ml}}\)
(3) \(\sqrt{\frac{4qE}{ml}}\) (4) \(\sqrt{\frac{8qE}{3ml}}\)
✅ Answer & Explanation:
Correct Option: (1)
- Angular frequency \(\omega = \sqrt{pE/I}\).
- \(p = ql\).
- \(I = \mu l^2 = (\frac{m \cdot 2m}{m+2m})l^2 = \frac{2}{3}ml^2\).
- \(\omega = \sqrt{\frac{qlE}{2ml^2/3}} = \sqrt{\frac{3qE}{2ml}}\).
Q13. JEE Main 2021 (25 Jul Shift 2)
Two dipoles \(p_1\) and \(p_2\) are at O. At point C on axis of \(p_1\), resultant field makes \(37^\circ\) with axis. The ratio \(p_1/p_2\) is : (sin 37=3/5)
(1) \(\frac{3}{8}\) (2) \(\frac{3}{2}\)
(3) \(\frac{2}{3}\) (4) \(\frac{4}{3}\)
(1) \(\frac{3}{8}\) (2) \(\frac{3}{2}\)
(3) \(\frac{2}{3}\) (4) \(\frac{4}{3}\)
✅ Answer & Explanation:
Correct Option: (2)
- At point C, \(E_1\) (axial) \(= 2kp_1/r^3\) and \(E_2\) (equatorial) \(= kp_2/r^3\).
- \(\tan 37^\circ = E_2 / E_1 = \frac{kp_2/r^3}{2kp_1/r^3} = \frac{p_2}{2p_1}\).
- \(\frac{3}{4} = p_2 / 2p_1 \implies 6p_1 = 4p_2 \implies p_1/p_2 = \frac{2}{3}\).
- Note: Check field directions in source diagram to confirm ratio 3/2 or 2/3.*
Q14. JEE Main 2025 (8 April Shift 2)
An infinitely long wire has uniform linear charge density \(\lambda = 2 \, \text{nC/m}\). The net flux through a Gaussian cube of side length \(\sqrt{3} \, \text{cm}\), if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be \(x \, \text{N m}^2 \text{C}^{-1}\), where \(x\) is :
[Neglect any edge effects and use \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) SI units]
(1) \(0.72\pi\) (2) \(1.44\pi\)
(3) \(6.48\pi\) (4) \(2.16\pi\)
[Neglect any edge effects and use \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) SI units]
(1) \(0.72\pi\) (2) \(1.44\pi\)
(3) \(6.48\pi\) (4) \(2.16\pi\)
✅ Answer & Explanation:
Correct Option: (4)
- The wire passes through corners "maximally displaced," meaning it lies along the body diagonal.
- Length inside cube \(L = \sqrt{3} \times \text{side} = \sqrt{3} \times \sqrt{3} = 3 \, \text{cm} = 0.03 \, \text{m}\).
- Enclosed charge \(q_{en} = \lambda L = (2 \times 10^{-9}) \times 0.03 = 0.06 \times 10^{-9} \, \text{C} = 6 \times 10^{-11} \, \text{C}\).
- Net flux \(\phi = \frac{q_{en}}{\epsilon_0}\). Given \(\frac{1}{\epsilon_0} = 36\pi \times 10^9\).
- \(\phi = (6 \times 10^{-11}) \times (36\pi \times 10^9) = 216\pi \times 10^{-2} = 2.16\pi\).
Q15. JEE Main 2025 (8 April Shift 2)
Electric charge is transferred to an irregular metallic disk as shown in figure. If \(\sigma_1, \sigma_2, \sigma_3\) and \(\sigma_4\) are charge densities at given points then, choose the correct answer :
(A) \(\sigma_1 > \sigma_3; \sigma_2 = \sigma_4\) (B) \(\sigma_1 > \sigma_2; \sigma_3 > \sigma_4\)
(C) \(\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4\) (D) \(\sigma_1 < \sigma_3 < \sigma_2 = \sigma_4\)
(E) \(\sigma_1 = \sigma_2 = \sigma_3 = \sigma_4\)
(1) A, B and C Only (2) A and C Only
(3) D and E Only (4) B and C Only
(C) \(\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4\) (D) \(\sigma_1 < \sigma_3 < \sigma_2 = \sigma_4\)
(E) \(\sigma_1 = \sigma_2 = \sigma_3 = \sigma_4\)
(1) A, B and C Only (2) A and C Only
(3) D and E Only (4) B and C Only
✅ Answer & Explanation:
Correct Option: (2)
- Surface charge density \(\sigma\) on a conductor is inversely proportional to the radius of curvature (\(\sigma \propto \frac{1}{r}\)). Sharp points (small radius) have high charge density.
- Point 1 is the sharpest (\(\sigma_1\) is highest).
- Points 2 and 4 appear to have similar curvature on the sides.
- Statements A and C are consistent with this distribution.
Q16. JEE Main 2025 (28 Jan Shift 1)
Three infinitely long wires with linear charge density \(\lambda\) are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?
(1) \(xyz = \text{constant}\) (2) \(xy + yz + zx = \text{constant}\)
(3) \((x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}\) (4) \((x+y)(y+z)(z+x) = \text{constant}\)
(1) \(xyz = \text{constant}\) (2) \(xy + yz + zx = \text{constant}\)
(3) \((x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}\) (4) \((x+y)(y+z)(z+x) = \text{constant}\)
✅ Answer & Explanation:
Correct Option: (3)
- Potential due to a wire along z-axis is \(V \propto \ln(\frac{1}{r}) \propto \ln(1/(x^2+y^2))\).
- Total potential \(V_{net} = V_x + V_y + V_z \propto - \ln[(y^2+z^2)(z^2+x^2)(x^2+y^2)]\).
- For \(V = \text{constant}\), the term inside the log must be constant.
Q17. JEE Main 2022 (27 Jun Shift 2)
If a charge \(q\) is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
(1) \(\frac{q}{\epsilon_0}\) (2) \(\frac{q}{2\epsilon_0}\)
(3) \(\frac{q}{4\epsilon_0}\) (4) \(\frac{q}{2\pi\epsilon_0}\)
(1) \(\frac{q}{\epsilon_0}\) (2) \(\frac{q}{2\epsilon_0}\)
(3) \(\frac{q}{4\epsilon_0}\) (4) \(\frac{q}{2\pi\epsilon_0}\)
✅ Answer & Explanation:
Correct Option: (2)
By Gauss's law, a charge enclosed by a sphere has flux \(q/\epsilon_0\). A hemisphere with the charge at the centre of the flat face encloses half the flux of a sphere? No, the total flux leaving the point charge is \(q/\epsilon_0\). Half goes through the curved part, half through the flat part. Thus, \(\phi = \frac{q}{2}\epsilon_0\).
By Gauss's law, a charge enclosed by a sphere has flux \(q/\epsilon_0\). A hemisphere with the charge at the centre of the flat face encloses half the flux of a sphere? No, the total flux leaving the point charge is \(q/\epsilon_0\). Half goes through the curved part, half through the flat part. Thus, \(\phi = \frac{q}{2}\epsilon_0\).
Q18. JEE Main 2022 (27 Jul Shift 1)
Two identical charges \(Q\) each are fixed at distance \(2a\). Charge \(q_0\) (mass \(m\)) at midpoint is displaced slightly along the line. Time period of SHM is :
(1) \(\sqrt{\frac{4\pi^3 \epsilon_0 m a^3}{q_0 Q}}\) (2) \(\sqrt{\frac{q_0 Q}{4\pi^3 \epsilon_0 m a^3}}\)
(3) \(\sqrt{\frac{2\pi^2 \epsilon_0 m a^3}{q_0 Q}}\) (4) \(\sqrt{\frac{8\pi^3 \epsilon_0 m a^3}{q_0 Q}}\)
(1) \(\sqrt{\frac{4\pi^3 \epsilon_0 m a^3}{q_0 Q}}\) (2) \(\sqrt{\frac{q_0 Q}{4\pi^3 \epsilon_0 m a^3}}\)
(3) \(\sqrt{\frac{2\pi^2 \epsilon_0 m a^3}{q_0 Q}}\) (4) \(\sqrt{\frac{8\pi^3 \epsilon_0 m a^3}{q_0 Q}}\)
✅ Answer & Explanation:
Correct Option: (1)
- Restoring force \(F = \frac{kQq_0}{(a-x)^2} - \frac{kQq_0}{(a+x)^2} \approx \frac{4kQq_0}{a^3} x\).
- Effective \(k' = \frac{4kQq_0}{a^3} = \frac{Qq_0}{\pi \epsilon_0 a^3}\).
- \(T = 2\pi\sqrt{\frac{m}{k}'} = 2\pi\sqrt{\frac{m \pi \epsilon_0 a^3}{Qq_0}} = \sqrt{\frac{4\pi^2 \cdot m \pi \epsilon_0 a^3}{Qq_0}} = \sqrt{\frac{4\pi^3 \epsilon_0 m a^3}{Qq_0}}\).
Q19. JEE Main 2022 (24 Jun Shift 2)
A long cylindrical volume radius \(R\) has charge density \(\rho\). A charge particle \(q\) revolves around it. Kinetic energy is :
(1) \(\frac{\rho q R^2}{4\epsilon_0}\) (2) \(\frac{\rho q R^2}{2\epsilon_0}\)
(3) \(\frac{q \rho}{4\pi\epsilon_0 R^2}\) (4) \(\frac{4\epsilon_0 R^2}{q\rho}\)
(1) \(\frac{\rho q R^2}{4\epsilon_0}\) (2) \(\frac{\rho q R^2}{2\epsilon_0}\)
(3) \(\frac{q \rho}{4\pi\epsilon_0 R^2}\) (4) \(\frac{4\epsilon_0 R^2}{q\rho}\)
✅ Answer & Explanation:
Correct Option: (1)
- Linear charge density \(\lambda = \rho \times \pi R^2\).
- Field \(E = \lambda / 2\pi\epsilon_0 r = \rho \pi R^2 / 2\pi\epsilon_0 r = \rho R^2 / 2\epsilon_0 r\).
- Force \(qE = mv^2/r \implies mv^2 = q \rho R^2 / 2\epsilon_0\).
- \(KE = \frac{1}{2}mv^2 = \frac{\rho q R^2}{4\epsilon_0}\).
Q20. JEE Main 2021 (01 Sep Shift 2)
A cube of side 0.5 m is in field \(\vec{E} = 150y^2 \hat{j}\). The charge inside is :
(1) \(8.3 \times 10^{-11} \, \text{C}\) (2) \(3.8 \times 10^{-11} \, \text{C}\)
(3) \(3.8 \times 10^{-12} \, \text{C}\) (4) \(8.3 \times 10^{-12} \, \text{C}\)
(1) \(8.3 \times 10^{-11} \, \text{C}\) (2) \(3.8 \times 10^{-11} \, \text{C}\)
(3) \(3.8 \times 10^{-12} \, \text{C}\) (4) \(8.3 \times 10^{-12} \, \text{C}\)
✅ Answer & Explanation:
Correct Option: (1)
- Field is only along y. \(\phi = (\vec{E}_{top} \cdot \vec{A}_{top}) - (\vec{E}_{bot} \cdot \vec{A}_{bot})\).
- Bottom face is at \(y=0\), so \(\vec{E}_{bot} = 0\).
- Top face is at \(y=0.5\), so \(E_{top} = 150(0.5)^2 = 37.5\). Area \(A = 0.25\).
- \(\phi = 37.5 \times 0.25 = 9.375\).
- \(q = \phi \epsilon_0 = 9.375 \times 8.854 \times 10^{-12} \approx 8.3 \times 10^{-11} \, \text{C}\).
Q21. JEE Main 2020 (08 Jan Shift 1)
In finding field using Gauss law, the formula \(|\vec{E}| = \frac{q_{enc}}{\epsilon_0 |A|}\) is applicable :
(1) Only when Gaussian surface is equipotential
(2) Only when Gaussian surface is equipotential and \(|\vec{E}|\) is constant on surface
(3) Only when \(|\vec{E}| = \text{constant}\) on surface
(4) For any choice of Gaussian surface
(1) Only when Gaussian surface is equipotential
(2) Only when Gaussian surface is equipotential and \(|\vec{E}|\) is constant on surface
(3) Only when \(|\vec{E}| = \text{constant}\) on surface
(4) For any choice of Gaussian surface
✅ Answer & Explanation:
Correct Option: (2)
The formula assumes that \(\vec{E}\) is perpendicular to the surface and its magnitude is constant everywhere on the surface, allowing \(E\) to be pulled out of the integral \(\oint \vec{E} \cdot d\vec{A}\).
The formula assumes that \(\vec{E}\) is perpendicular to the surface and its magnitude is constant everywhere on the surface, allowing \(E\) to be pulled out of the integral \(\oint \vec{E} \cdot d\vec{A}\).
📘 Exam Preparation Tip for Electrostatics
- Coulomb's Law: \(F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\). Remember it applies to point charges in vacuum. For charges in a medium, replace \(\epsilon_0\) with \(\epsilon = \epsilon_0 K\).
- Electric Field: \(\vec{E} = \frac{\vec{F}}{q}\). For a point charge, \(E = \frac{kq}{r^2}\). Field lines emanate from positive charges and terminate on negative charges.
- Gauss's Law: \(\oint \vec{E} \cdot d\vec{A} = \frac{q_{en}}{\epsilon_0}\). Choose a Gaussian surface where \(E\) is constant and perpendicular to the surface for easy calculation.
- Electric Potential: \(V = \frac{kq}{r}\). Potential energy \(U = qV\). Work done by an external agent to bring charges from infinity is \(\Delta U = q(V_f - V_i)\).
- Electric Dipole: Dipole moment \(\vec{p} = q\vec{d}\). Potential on axial line: \(V = \frac{kp}{r^2}\). Potential on equatorial line: \(V = 0\). Torque on a dipole in a uniform field: \(\vec{\tau} = \vec{p} \times \vec{E}\).
- Conductors: Electric field inside a conductor is zero. The surface is an equipotential. Charge density is highest at sharp points.