Question

An air capacitor of capacitance 0.005 μF is connected to a direct voltage of 500 V. It is disconnected and then immersed in oil with a relative permittivity of 2.5. The energy after immersion is?

(a) 275 μJ

(b) 250 μJ

(c) 225 μJ

(d) 625 μJ

This question was posed to me during an internship interview.

Asked question is from Problems Involving Dot Conventions topic in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer 1

Right option is (b) 250 μJ

The explanation: E = \(\frac{1}{2}\) CV^2

Or, C = \(\frac{ε_0 ε_r A}{d} = ϵ_r (\frac{ε_0 A}{d})\)

= 2.5 × 0.005 × 10^-6

∴ CNEW = 12.5 × 10^-9 F

Now, q = CV = 0.005 × 10^-6 × 500 = 2.5 × 10^-6

VNEW = \(\frac{q} {C_{NEW}}\)

= \(\frac{2.5 × 10^{-6}}{12.5 × 10^{-9}}\) VNEW = 200

E = \(\frac{1}{2}\) CV^2

= \(\frac{1}{2}\) × 12.5 × 10^-9 × (200)^2

= 250 μJ.