Exams › JEE Main › Physics

A parallel plate capacitor with area 200 cm² and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is 25 × 10⁻⁶ N, the value of V is approximately. ε₀ = 8.85 × 10⁻¹² C²/N·m²

1. 150 V
2. 100 V
3. 250 V
4. 300 V

Correct answer: 250 V

Solution

Force between plates F = eps0 A V^2 / (2 d^2), so V = sqrt(2 F d^2 / (eps0 A)). With A = 200 cm^2 = 0.02 m^2, d = 0.015 m, F = 25x10^-6 N: V = sqrt(2 x 25e-6 x 0.015^2 / (8.85e-12 x 0.02)) ~ 252 V, i.e. about 250 V.

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 →