Question

A parallel plate capacitor has area of each plate A, the separation between the plates d. It is charged to a potential V and then disconnected from the battery. How much work will be done in filling the capacitor completely with a dielectric of constant K?

• A. $\frac{1}{2} \frac{\epsilon_0 AV^2}{d} (1 - \frac{1}{K^2})$
• B. $\frac{1}{2} \frac{V^2 \epsilon_0 A}{Kd}$
• C. $\frac{1}{2} \frac{V^2 \epsilon_0 A}{K^2d}$
• D. $\frac{1}{2} \frac{\epsilon_0 AV^2}{d} (1 - \frac{1}{K})$

Answer 1

Answer: (D)

(4)

Answer 2

The problem asks for the work done in filling a parallel plate capacitor with a dielectric after it has been charged and disconnected from the battery. When the capacitor is disconnected from the battery, the charge on its plates remains constant.

1. Initial State (before filling with dielectric):

• Area of each plate = $A$
• Separation between plates = $d$
• Initial capacitance, $C_0 = \frac{\epsilon_0 A}{d}$
• The capacitor is charged to a potential $V$. So, the initial potential difference across the plates is $V_0 = V$.
• The charge stored on the plates, $Q = C_0 V_0 = \frac{\epsilon_0 A V}{d}$.
• The initial energy stored in the capacitor, $U_0 = \frac{1}{2} C_0 V_0^2 = \frac{1}{2} \left(\frac{\epsilon_0 A}{d}\right) V^2$.

2. After Disconnection from Battery:

• Once disconnected, the charge $Q$ on the plates remains constant.

3. Final State (after filling completely with dielectric):

• The dielectric constant of the material is $K$.
• The new capacitance, $C_f = K C_0 = K \frac{\epsilon_0 A}{d}$.
• Since the charge $Q$ is constant, the new potential difference across the plates will be $V_f = \frac{Q}{C_f}$. Substituting $Q = C_0 V_0$ and $C_f = K C_0$: $V_f = \frac{C_0 V_0}{K C_0} = \frac{V_0}{K} = \frac{V}{K}$.
• The final energy stored in the capacitor can be calculated using the constant charge $Q$: $U_f = \frac{1}{2} \frac{Q^2}{C_f} = \frac{1}{2} \frac{Q^2}{K C_0}$. We know $U_0 = \frac{1}{2} \frac{Q^2}{C_0}$. Therefore, $U_f = \frac{U_0}{K}$. Substituting the expression for $U_0$: $U_f = \frac{1}{K} \left( \frac{1}{2} \frac{\epsilon_0 A V^2}{d} \right) = \frac{1}{2} \frac{\epsilon_0 A V^2}{Kd}$.

4. Work Done:

When a dielectric is inserted into a capacitor with constant charge, the energy stored in the capacitor decreases ($U_f < U_0$ since $K>1$). This decrease in energy is released as work done by the electric field on the dielectric, pulling it into the capacitor. The work done in filling the capacitor is the energy released, which is the difference between the initial and final stored energies: Work Done ($W$) = $U_0 - U_f$ $W = \frac{1}{2} \frac{\epsilon_0 A V^2}{d} - \frac{1}{2} \frac{\epsilon_0 A V^2}{Kd}$ Factor out the common term: $W = \frac{1}{2} \frac{\epsilon_0 A V^2}{d} \left( 1 - \frac{1}{K} \right)$

The calculated work done matches option (4).