Practice portal › Electric Potential and Capacitance › Capacitor Circuits and Energy

A 2 μ F capacitor C₁ is first charged to a potential difference of 10 V using a battery through a switch S₁. Then the battery is removed and a second switch S₂ connects the capacitor across an uncharged capacitor C₂ of 8 μ F. The charge in C₂ at the equilibrium condition is ______ μ C. (Round off to the nearest integer.)

Asked in JEE Main 17th March 2nd Shift 2021 · Redistribution of charge

Figure: Redistribution of charge
Answer: 16

Step-by-step solution

Charge placed on C₁: Q=C₁V=2×10=20 μ C.

With the battery removed, this charge is conserved and redistributes until both capacitors sit at a common potential.

V_common=Q/(C₁+C₂)=(20)/(2+8)=2 V.

q₂=C₂V_common=8×2=16 μ C.

More Capacitor Circuits and Energy questionsAll Capacitor Circuits and Energy questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer