The JEE Main Physics section requires speed and accuracy, along with a thorough understanding of the Capacitance. This article provides a set of JEE Main PYQs on Capacitance to help you understand the topic and improve your problem-solving skills with the help of detailed solutions by ensuring conceptual clarity, which will help you in the JEE Main 2026 preparation.
Whether you're revising the basics or testing your knowledge, these JEE Main PYQs will serve as a valuable practice resource.
The JEE Main 2026 exam is likely to continue on the same pattern as JEE Main 2025. Out of 90 questions, students can expect a fair mix from all three subjects. To get an edge, going through JEE Main previous year questions (PYQs) is one of the best strategies.
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JEE Main PYQs on Capacitance
1.
A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure. If the area of each stair is \(\frac{A}{3}\) and the height is d, the capacitance of the arrangement is:
- \( \frac{11 \varepsilon_0 A}{18 d} \)
- \( \frac{13 \varepsilon_0 A}{17 d} \)
- \( \frac{11 \varepsilon_0 A}{20 d} \)
- \( \frac{18 \varepsilon_0 A}{11 d} \)
2.
A parallel plate capacitor was made with two rectangular plates, each with a length of \( l = 3 \, {cm} \) and breadth of \( b = 1 \, {cm} \). The distance between the plates is \( d = 3 \, \mu{m} \). Out of the following, which are the ways to increase the capacitance by a factor of 10?- A. l = 30 cm, b = 1 cm, d = 1 μm
- B. l = 3 cm, b = 1 cm, d = 30 μm
- C. l = 6 cm, b = 5 cm, d = 3 μm
- D. l = 1 cm, b = 1 cm, d = 10 μm
- E. l = 5 cm, b = 2 cm, d = 1 μm
- C and E only
- B and D only
- A only
- C only
3.
A parallel plate capacitor of capacitance 1 μF is charged to a potential difference of 20 V. The distance between plates is 1 μm. The energy density between the plates of the capacitor is:- \( 2 \times 10^{-4} \, {J/m}^3 \)
- \( 1.8 \times 10^5 \, {J/m}^3 \)
- \( 1.8 \times 10^3 \, {J/m}^3 \)
- \( 2 \times 10^2 \, {J/m}^3 \)
4.
Consider a parallel plate capacitor of area \( A \) (of each plate) and separation \( d \) between the plates. If \( E \) is the electric field and \( \epsilon_0 \) is the permittivity of free space between the plates, then the potential energy stored in the capacitor is:- \( \frac{1}{2} \epsilon_0 E^2 A d \)
- \( \frac{3}{4} \epsilon_0 E^2 A d \)
- \( \frac{1}{4} \epsilon_0 E^2 A d \)
- \( \epsilon_0 E^2 A d \)
5.
A galvanometer (G) of $2 \, \Omega$ resistance is connected in the given circuit. The ratio of charge stored in $C_1$ and $C_2$ is:
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
- 1
- \(\frac{1}{2}\)
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