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📘 Capacitance — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Capacitance for JEE Main with previous year questions on capacitors, dielectrics, and RC circuits. Each solution is explained in a simple, step-by-step manner to boost your score.
Capacitance — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (8 April Shift 2)
Space between the plates of a parallel plate capacitor of plate area \(4 \, \text{cm}^2\) and separation of (d) 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is ________ pF. (Given \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\))
✅ Answer & Explanation:
Correct Answer: 15
- The two dielectric layers act as two capacitors in series:
- \(C_1 = \frac{k_1 \epsilon_0 A}{\frac{d}{2}} = \frac{5 \epsilon_0 A}{\frac{d}{2}} = \frac{10 \epsilon_0 A}{d}\)
- \(C_2 = \frac{k_2 \epsilon_0 A}{\frac{d}{2}} = \frac{3 \epsilon_0 A}{\frac{d}{2}} = \frac{6 \epsilon_0 A}{d}\)
- Series combination \(C_s = \frac{C_1 C_2}{C_1 + C_2} = \frac{60}{16} \frac{\epsilon_0 A}{d} = 3.75 \frac{\epsilon_0 A}{d}\).
- Substituting values: \(C_s = 3.75 \times \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}} = 3.75 \times 2 \times 10^{-12} = 7.5 \, \text{pF}\).
- Total effective capacitance (parallel) \(= 7.5 + 7.5 = 15 \, \text{pF}\).
Q2. JEE Main 2025 (4 April Shift 1)
Four capacitor each of capacitance \(16 \, \mu\text{F}\) are connected as shown in the figure. The capacitance between points A and B is : ______ \(\mu\text{F}\).
✅ Answer & Explanation:
Correct Answer: 64
- Analysis of the junction points reveals that all four capacitors are connected in parallel between points A and B.
- Therefore, \(C_{eq} = C_1 + C_2 + C_3 + C_4 = 4 \times 16 = 64 \, \mu\text{F}\).
Q3. JEE Main 2025 (24 Jan Shift 1)
A parallel plate capacitor was made with two rectangular plates, each with a length of \(l = 3 \, \text{cm}\) and breath of \(b = 1 \, \text{cm}\). The distance between the plates is \(3 \, \mu\text{m}\). Out of the following, which are the ways to increase the capacitance by a factor of 10 ?
A. \(l = 30 \, \text{cm}, b = 1 \, \text{cm}, d = 1 \, \mu\text{m}\)
B. \(l = 3 \, \text{cm}, b = 1 \, \text{cm}, d = 30 \, \mu\text{m}\)
C. \(l = 6 \, \text{cm}, b = 5 \, \text{cm}, d = 3 \, \mu\text{m}\)
D. \(l = 1 \, \text{cm}, b = 1 \, \text{cm}, d = 10 \, \mu\text{m}\)
E. \(l = 5 \, \text{cm}, b = 2 \, \text{cm}, d = 1 \, \mu\text{m}\)
(1) A only (2) C only (3) B and D only (4) C and E only
A. \(l = 30 \, \text{cm}, b = 1 \, \text{cm}, d = 1 \, \mu\text{m}\)
B. \(l = 3 \, \text{cm}, b = 1 \, \text{cm}, d = 30 \, \mu\text{m}\)
C. \(l = 6 \, \text{cm}, b = 5 \, \text{cm}, d = 3 \, \mu\text{m}\)
D. \(l = 1 \, \text{cm}, b = 1 \, \text{cm}, d = 10 \, \mu\text{m}\)
E. \(l = 5 \, \text{cm}, b = 2 \, \text{cm}, d = 1 \, \mu\text{m}\)
(1) A only (2) C only (3) B and D only (4) C and E only
✅ Answer & Explanation:
Correct Option: (4)
- Original \(C \propto (3 \times 1) / 3 = 1\).
- (C): New \(C \propto (6 \times 5) / 3 = 10\). (Factor of 10)
- (E): New \(C \propto (5 \times 2) / 1 = 10\). (Factor of 10)
Q4. JEE Main 2022 (28 Jul Shift 1)
Two capacitors, each having capacitance \(40 \, \mu\text{F}\) are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became \(24 \, \mu\text{F}\). The value of K will be :
(1) 1.5 (2) 2.5
(3) 1.2 (4) 3
(1) 1.5 (2) 2.5
(3) 1.2 (4) 3
✅ Answer & Explanation:
Correct Option: (1)
- \(\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{KC} \implies \frac{1}{24} = \frac{1}{40}(1 + \frac{1}{K})\).
- \(\frac{40}{24} = 1 + \frac{1}{K} \implies \frac{5}{3} - 1 = \frac{1}{K} \implies \frac{2}{3} = \frac{1}{K} \implies K = 1.5\).
Q5. JEE Main 2022 (27 Jun Shift 1)
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be:
(1) 5 N (2) 10 N
(3) 20 N (4) Zero
(1) 5 N (2) 10 N
(3) 20 N (4) Zero
✅ Answer & Explanation:
Correct Option: (1)
- Field between two plates is \(E = \sigma / \epsilon_0\).
- Field due to a single plate is \(E' = \sigma / 2\epsilon_0 = \frac{E}{2}\).
- Since \(F = qE\), then \(F' = \frac{F}{2} = 5 \, \text{N}\).
Q6. JEE Main 2022 (25 Jul Shift 1)
A condenser of \(2 \, \mu\text{F}\) capacitance is charged steadily from 0 to 5 C. Which of the following graph represents correctly the variation of potential difference (V) across its plates with respect to the charge (Q) on the condenser?
Choose correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (1)
Since \(Q = CV\), then \(V = (\frac{1}{C}) Q\). This is a linear relationship passing through the origin with a positive slope. Graph (1) correctly represents this.
Since \(Q = CV\), then \(V = (\frac{1}{C}) Q\). This is a linear relationship passing through the origin with a positive slope. Graph (1) correctly represents this.
Q7. JEE Main 2020 (08 Jan Shift 2)
A capacitor is made of two square plates each of side ‘a’ making a very small angle \(\alpha\) between them, as shown in figure. The capacitance will be close to:
(1) \(\frac{\epsilon_0 a^2}{d} (1 - \frac{\alpha a}{2d})\) (2) \(\frac{\epsilon_0 a^2}{d} (1 - \frac{\alpha a}{4d})\)
(3) \(\frac{\epsilon_0 a^2}{d} (1 + \frac{\alpha a}{d})\) (4) \(\frac{\epsilon_0 a^2}{d} (1 - \frac{3\alpha a}{2d})\)
(1) \(\frac{\epsilon_0 a^2}{d} (1 - \frac{\alpha a}{2d})\) (2) \(\frac{\epsilon_0 a^2}{d} (1 - \frac{\alpha a}{4d})\)
(3) \(\frac{\epsilon_0 a^2}{d} (1 + \frac{\alpha a}{d})\) (4) \(\frac{\epsilon_0 a^2}{d} (1 - \frac{3\alpha a}{2d})\)
✅ Answer & Explanation:
Correct Option: (1)
- For small \(\alpha\), the distance between plates varies as \(d(x) = d + \alpha x\).
- Integrating small strips of capacitance \(dC = \frac{\epsilon_0 a dx}{d + \alpha x}\) from \(x=0\) to \(a\) yields the approximate expression in Option (1).
Q8. JEE Main 2025 (28 Jan Shift 2)
A parallel plate capacitor of capacitance \(1 \, \mu\text{F}\) is charged to a potential difference of 20 V. The distance between plates is \(1 \, \mu\text{m}\). The energy density between plates of capacitor is.
(1) \(2 \times 10^{-4} \, \text{J/m}^3\) (2) \(1.8 \times 10^5 \, \text{J/m}^3\)
(3) \(1.8 \times 10^3 \, \text{J/m}^3\) (4) \(2 \times 10^2 \, \text{J/m}^3\)
(1) \(2 \times 10^{-4} \, \text{J/m}^3\) (2) \(1.8 \times 10^5 \, \text{J/m}^3\)
(3) \(1.8 \times 10^3 \, \text{J/m}^3\) (4) \(2 \times 10^2 \, \text{J/m}^3\)
✅ Answer & Explanation:
Correct Option: (3)
- Energy density \(u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 (\frac{V}{d})^2\).
- Using standard \(\epsilon_0 \approx 9 \times 10^{-12}\):
- \(u = 0.5 \times (8.85 \times 10^{-12}) \times (20 / 10^{-6})^2 \approx 1.8 \times 10^3 \, \text{J/m}^3\).
Q9. JEE Main 2025 (23 Jan Shift 1)
Identify the valid statements relevant to the given circuit at the instant when the key is closed.
A. There will be no current through resistor R.
B. There will be maximum current in the connecting wires.
C. Potential difference between the capacitor plates A and B is minimum.
D. Charge on the capacitor plates is minimum.
(1) A, C Only (2) A, B, D Only (3) C, D Only (4) B, C, D Only
B. There will be maximum current in the connecting wires.
C. Potential difference between the capacitor plates A and B is minimum.
D. Charge on the capacitor plates is minimum.
(1) A, C Only (2) A, B, D Only (3) C, D Only (4) B, C, D Only
✅ Answer & Explanation:
Correct Option: (4)
- At \(t = 0\) (instant key is closed):
- Capacitor acts as a short circuit (zero resistance).
- Current is maximum (\(I = \frac{V}{R}\)). [B is true]
- Charge and potential across capacitor are zero (minimum). [C and D are true]
Q10. JEE Main 2025 (22 Jan Shift 1)
A parallel-plate capacitor of capacitance \(40 \, \mu\text{F}\) is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant \(K = 2\). Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are:
(1) 4 mC and 0.2 J (2) 8 mC and 2.0 J
(3) 2 mC and 0.4 J (4) 2 mC and 0.2 J
(1) 4 mC and 0.2 J (2) 8 mC and 2.0 J
(3) 2 mC and 0.4 J (4) 2 mC and 0.2 J
✅ Answer & Explanation:
Correct Option: (1)
- Initial \(Q_i = CV = 40 \times 10^{-6} \times 100 = 4 \, \text{mC}\). Energy \(U_i = 0.5 CV^2 = 0.2 \, \text{J}\).
- Final \(Q_f = (KC)V = 2 \times 4 \, \text{mC} = 8 \, \text{mC}\). Energy \(U_f = 0.5 (KC)V^2 = 2 \times 0.2 \, \text{J} = 0.4 \, \text{J}\).
- Extra charge \(\Delta Q = 4 \, \text{mC}\). Change in energy \(\Delta U = 0.2 \, \text{J}\).
Q11. JEE Main 2024 (05 Apr Shift 1)
The electric field between the two parallel plates of a capacitor of \(1.5 \, \mu\text{F}\) capacitance drops to one third of its initial value in \(6.6 \, \mu\text{s}\) when the plates are connected by a thin wire. The resistance of this wire is _____ \(\Omega\). (Given, \(\log 3 = 1.1\))
✅ Answer & Explanation:
Correct Answer: 4
- Discharge equation: \(E = E_0 e^{-t/RC}\).
- \(\frac{1}{3} = e^{-6.6 / (R \times 1.5)}\)
- \(\ln 3 = 6.6 / 1.5R \implies 2.303 \times 1.1 = 6.6 / 1.5R \implies 2.53 \approx 4.4 / R \implies R \approx 4 \, \Omega\).
Q12. JEE Main 2024 (04 Apr Shift 2)
A parallel plate capacitor of capacitance 12.5 pF is charged by a battery to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab (\(\epsilon_r = 6\)) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _____ \(10^{-12} \, \text{J}\).
✅ Answer & Explanation:
Correct Answer: 750
- Initial energy \(U_i = \frac{1}{2} CV^2 = 0.5 \times 12.5 \times (12)^2 = 6.25 \times 144 = 900 \, \text{pJ}\).
- Battery disconnected, charge \(Q\) is constant. Final energy \(U_f = U_i / K = 900 / 6 = 150 \, \text{pJ}\).
- Change \(|\Delta U| = 900 - 150 = 750 \, \text{pJ} = 750 \times 10^{-12} \, \text{J}\).
Q13. JEE Main 2025 (29 Jan Shift 2)
A capacitor, \(C_1 = 6 \, \mu\text{F}\) is charged to a potential difference of \(V_0 = 5 \, \text{V}\) using a 5 V battery. The battery is removed and another capacitor, \(C_2 = 12 \, \mu\text{F}\) is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges (\(q_1\) and \(q_2\)) on the capacitors \(C_1\) and \(C_2\) when equilibrium condition is reached.
(1) \(q_1 = 10 \, \mu\text{C}, q_2 = 20 \, \mu\text{C}\) (2) \(q_1 = 30 \, \mu\text{C}, q_2 = 15 \, \mu\text{C}\)
(3) \(q_1 = 20 \, \mu\text{C}, q_2 = 10 \, \mu\text{C}\) (4) \(q_1 = 15 \, \mu\text{C}, q_2 = 30 \, \mu\text{C}\)
(3) \(q_1 = 20 \, \mu\text{C}, q_2 = 10 \, \mu\text{C}\) (4) \(q_1 = 15 \, \mu\text{C}, q_2 = 30 \, \mu\text{C}\)
✅ Answer & Explanation:
Correct Option: (1)
- Initial total charge \(Q = C_1 V_0 = 6 \times 5 = 30 \, \mu\text{C}\).
- Common potential \(V = \frac{Q}{C_1 + C_2} = \frac{30}{6 + 12} = \frac{30}{18} = \frac{5}{3} \, \text{V}\).
- \(q_1 = C_1 V = 6 \times (\frac{5}{3}) = 10 \, \mu\text{C}\).
- \(q_2 = C_2 V = 12 \times (\frac{5}{3}) = 20 \, \mu\text{C}\).
Q14. JEE Main 2025 (28 Jan Shift 1)
Two capacitors \(C_1\) and \(C_2\) are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are \(U_1\) and \(U_2\), respectively. Which of the given statements is true?
(1) \(C_2 > C_1, U_2 < U_1\) (2) \(C_1 > C_2, U_1 > U_2\)
(3) \(C_1 > C_2, U_1 < U_2\) (4) \(C_2 > C_1, U_2 > U_1\)
(3) \(C_1 > C_2, U_1 < U_2\) (4) \(C_2 > C_1, U_2 > U_1\)
✅ Answer & Explanation:
Correct Option: (4)
- In parallel, voltage \(V\) is same. Steady-state charge \(Q = CV\).
- From graph, steady-state charge for \(C_2\) is higher than \(C_1\), so \(C_2 > C_1\).
- Stored energy \(U = \frac{1}{2} CV^2 \propto C\). Since \(C_2 > C_1\), then \(U_2 > U_1\).
Q15. JEE Main 2024 (29 Jan Shift 1)
A \(16 \, \Omega\) wire is bend to form a square loop. A 9 V battery with internal resistance \(1 \, \Omega\) is connected across one of its sides. If a \(4 \, \mu\text{F}\) capacitor is connected across one of its diagonals, the energy stored by the capacitor will be \(\frac{x}{2} \, \mu\text{J}\), where \(x = \) ______.
✅ Answer & Explanation:
Correct Answer: 81
- Each side of the square has resistance \(R = 4 \, \Omega\).
- Diagonal potential difference \(\Delta V = 4.5 \, \text{V}\) (using bridge/voltage divider analysis).
- Energy \(U = \frac{1}{2} C (\Delta V)^2 = \frac{1}{2} (4 \times 10^{-6}) (4.5)^2 = 2 \times 20.25 = 40.5 \, \mu\text{J}\).
- \(40.5 = \frac{x}{2} \implies x = 81\).
Q16. JEE Main 2023 (13 Apr Shift 2)
In the circuit shown, the energy stored in the capacitor is \(n \, \mu\text{J}\). The value of \(n\) is _____.
✅ Answer & Explanation:
Correct Answer: 75
- Using Kirchhoff's laws for steady state (capacitor branch is open):
- Potential difference across the \(6 \, \mu\text{F}\) capacitor is found to be 5 V.
- \(U = \frac{1}{2} CV^2 = \frac{1}{2} (6) (5)^2 = 75 \, \mu\text{J}\). Thus \(n = 75\).
Q17. JEE Main 2022 (26 Jul Shift 1)
A composite parallel plate capacitor is made up of two different dielectric materials with different thickness (\(t_1 = 0.5 \, \text{mm}\) and \(t_2 = 1 \, \text{mm}\)) as shown in figure. The two different dielectric material are separated by a conducting foil F. The voltage of the conducting foil is _____ V.
✅ Answer & Explanation:
Correct Answer: 60
- The system is two capacitors in series. Let \(C_1 = k_1 A / t_1\) and \(C_2 = k_2 A / t_2\).
- Voltage division in series: \(V_2 = V_{total} \times \frac{C_1}{C_1 + C_2}\).
- Substituting values from the diagram results in a foil voltage of 60 V.
Q18. JEE Main 2022 (25 Jun Shift 1)
The equivalent capacitance between points A and B in below shown figure will be _____ \(\mu\text{F}\).
✅ Answer & Explanation:
Correct Answer: 6
Simplifying the series and parallel groups of the \(8 \, \mu\text{F}\) capacitors step-by-step reduces the network to an equivalent capacitance of 6 \(\mu\text{F}\).
Simplifying the series and parallel groups of the \(8 \, \mu\text{F}\) capacitors step-by-step reduces the network to an equivalent capacitance of 6 \(\mu\text{F}\).
Q19. JEE Main 2021 (27 Aug Shift 1)
Calculate the amount of charge on capacitor of \(4 \, \mu\text{F}\). The internal resistance of battery is \(1 \, \Omega\).
(1) \(4 \, \mu\text{C}\) (2) \(8 \, \mu\text{C}\)
(3) \(16 \, \mu\text{C}\) (4) Zero
(3) \(16 \, \mu\text{C}\) (4) Zero
✅ Answer & Explanation:
Correct Option: (2)
- In steady state, the capacitor branch draws no current. The current in the rest of the circuit is \(I = 5 / (4 + 1) = 1 \, \text{A}\).
- The potential difference across the capacitor branch is the voltage across the \(4 \, \Omega\) resistor: \(V = IR = 1 \times 2 = 2 \, \text{V}\).
- Charge \(Q = CV = 4 \times 2 = 8 \, \mu\text{C}\).
📘 Exam Preparation Tip for Capacitance
- Parallel Plate Capacitor: \(C = \frac{\epsilon_0 A}{d}\). The capacitance depends on the geometry (area and separation) and the medium between the plates.
- Dielectric Materials: When a dielectric of constant \(K\) is inserted, capacitance increases by a factor of \(K\): \(C = K C_0\). The electric field between the plates decreases by a factor of \(K\).
- Energy Stored: \(U = \frac{1}{2} CV^2 = \frac{Q^2}{2C} = \frac{1}{2} QV\). Energy density \(u = \frac{1}{2} \epsilon_0 E^2\).
- Series and Parallel Combinations: In parallel, \(C_{eq} = C_1 + C_2\). In series, \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\). For series, the charge is the same on each capacitor.
- Charge Sharing: When two capacitors are connected, charge redistributes until the potential difference is the same. The common potential is \(V = \frac{Q_1 + Q_2}{C_1 + C_2}\).
- RC Circuits: Charging: \(V(t) = V_0(1 - e^{-t/RC})\). Discharging: \(V(t) = V_0 e^{-t/RC}\). The time constant \(\tau = RC\) is the time for the voltage to rise to 63.2% of its final value (or fall to 36.8% of its initial value).