Answer to Question #131015 in Electric Circuits for NILAKSHA GHOSH

Question

A 50pF parallel-plate capacitor with an air gap between the plates is connected to a 50V battery. A Teflon(k=2)slab is then inserted between the plates and completely fills the gap.Then the battery is disconnected. What is the new potential difference?

Answer 1

Explanations & Calculations

• Capacitance ($\small C_0$ ) of a parallel plate capacitor with an air gap (without any dielectric medium hence permittivity is the absolute : $\epsilon_0$) is given by, $\small C_0 = \Large\frac{\epsilon_0A}{d}$
• When a dielectric medium is inserted (here completely fills the gap: d) only the permittivity is improved & increased (here the new permittivity is $\epsilon_1 = k\epsilon_0 = 2\epsilon_0$) hence the capacitance is increased (2$\small C_0$ ).
• According to $\small Q=CV$, amount of electrostatic charges the capacitor can bare is increased.
• Now what happens here is as the battery remains connected, the new arrangement gets charged until the potential difference equals the applied value (V) while baring more electrostatic charges inside.
• Therefore, when the battery is removed, potential difference remains the same.