Question

A parallel plate capacitor of capacitance \( 12.5 \, \text{pF} \) is charged by a battery connected between its plates to a potential difference of \( 12.0 \, \text{V} \). The battery is now disconnected and a dielectric slab (\( \varepsilon_r = 6 \)) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ______ \( \times 10^{-12} \, \text{J} \).

Answer 1

Correct Answer: 750

1. Initial Capacitance and Charge on Capacitor:
The initial capacitance \( C_0 = 12.5 \, \text{pF} \), and the initial charge on the capacitor \( Q = C_0V \).
2. Capacitance with Dielectric Inserted:
After inserting a dielectric with dielectric constant \( \epsilon_r = 6 \), the new capacitance becomes:
\[ C_f = \epsilon_r C_0. \]

3. Change in Potential Energy:
The change in potential energy of the capacitor is given by:
\[ \Delta U = U_i - U_f = \frac{Q^2}{2C_i} - \frac{Q^2}{2C_f}. \] Substituting \( Q = C_0V \) and simplifying:
\[ \Delta U = \frac{(C_0V)^2}{2C_0} \left[ 1 - \frac{1}{\epsilon_r} \right] = \frac{1}{2} C_0V^2 \left[ 1 - \frac{1}{6} \right]. \] 

4. Calculation:
Substitute \( C_0 = 12.5 \, \text{pF}, \, V = 12 \, \text{V}, \, \text{and} \, \epsilon_r = 6 \):
\[ \Delta U = \frac{1}{2} \times 12.5 \times 10^{-12} \times (12)^2 \times \frac{5}{6}. \] Simplifying further:
\[ \Delta U = 750 \, \text{pJ} = 750 \times 10^{-12} \, \text{J}. \] 

Answer: \( 750 \times 10^{-12} \, \text{J} \)