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

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: $750 \times 10^{-12} \, \text{J}$