Exams › JEE Main › Physics

A 10 μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:

1. 10 μF
2. 15 μF
3. 20 μF
4. 30 μF

Correct answer: 15 μF

Solution

Q = 10uF * 50V = 500uC. After parallel connection 20 = 500/(10 + C2), so 10 + C2 = 25 -> C2 = 15uF.

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 →