Exams › JEE Main › Physics

A capacitor with capacitance 5 μF is charged to 5 μC. If the plates are pulled apart to reduce the capacitance to 2 μF, how much work is done? [2020]

1. 6.25 × 10⁻⁶ J
2. 3.75 × 10⁻⁶ J
3. 2.16 × 10⁻⁶ J
4. 2.55 × 10⁻⁶ J

Correct answer: 3.75 × 10⁻⁶ J

Solution

The work done when a capacitor's capacitance changes while maintaining a constant charge can be calculated using the formula W = Q²/(2C). In this case, with an initial charge of 5 μC and a final capacitance of 2 μF, the work done is 3.75 × 10⁻⁶ J, confirming option B as correct.

Related JEE Main Physics questions

• If an electric field vector E exists in a material whose dielectric constant is K, and ε0 denotes the permittivity of free space, which expression gives the electric displacement vector?
• If a unit positive charge is taken from point A to point B along the same equipotential surface, then what can be said about the potential difference V_A - V_B?
• Two thin metallic spherical shells are placed concentrically with radii R1 and R2, where R1 > R2. If the shells carry charges Q1 and Q2 respectively, what is the electric potential at a point located at distance r such that R2 < r < R1? (Take k = 1/(4πϵ0))
• An alpha particle is made to move through a potential difference of 10⁶ V. What kinetic energy does it gain?
• A parallel-plate capacitor has its entire gap filled with a dielectric whose dielectric constant changes with position x according to K(x) = K0 + λx, where λ is a constant. How is its capacitance C related to the vacuum capacitance C0?
• Two thin circular rings of radius R each are kept on the same axis with their centres separated by a distance d. One ring carries charge +q and the other carries charge −q. What is the potential difference between the centres of the two rings?

⚔️ Practice JEE Main Physics free + battle 1v1 →