Question

A 193 volt battery is connected to a 0.3 microfarad parallel-plate capacitor with air between the plates. Now the battery is disconnected, with care taken not to discharge the capacitor. Some paper is then inserted between the plates completely filling up the space. What is the final potential difference between the plates ? (the dielectric constant for air is = 1 and for the paper inserted is = 3)

Answer 1

V = 193 Volt
C = 0.3 * 10^-6 F

Charge Stored, Q = C*V
Q = (0.3 * 10^-6 * 193) C
Q = 57.9 uC

Now Charge remains same, but Capacitance changes.
C = K*A*eo/d
C = 3.0 * 0.3 uF
C = 0.9 uF

Q = C * V
V = Q/C
V = 57.9/0.9
V = 64.3 V
Final potential difference between the plates, V = 64.3 V