Question

A parallel plate capacitor of capacitance 200 \(\mu F\) is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remains connected. The change in the electrostatic energy in the capacitor will be ________ J.

Hint:

Distinguish between "Battery Connected" (\(V\) constant) and "Battery Disconnected" (\(Q\) constant). If the battery were disconnected, the energy would have decreased by a factor of \(K\).

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
When a dielectric is inserted into a capacitor while the battery is still connected, the potential difference (\(V\)) remains constant, but the capacitance increases. This leads to an increase in the stored energy.
Step 2: Key Formula or Approach:
1. Initial Energy: \( U_i = \frac{1}{2} C V^2 \).
2. Final Energy: \( U_f = \frac{1}{2} C' V^2 = \frac{1}{2} (KC) V^2 \).
3. Change in Energy: \( \Delta U = U_f - U_i = \frac{1}{2} (K-1) C V^2 \).
Step 3: Detailed Explanation:
Given: \( C = 200 \mu F = 2 \times 10^{-4} \text{ F} \), \( V = 200 \text{ V} \), \( K = 2 \).
\[ \Delta U = \frac{1}{2} (2-1) \times (2 \times 10^{-4}) \times (200)^2 \]
\[ \Delta U = \frac{1}{2} \times 1 \times 2 \times 10^{-4} \times 40000 \]
\[ \Delta U = 1 \times 10^{-4} \times 4 \times 10^4 = 4 \text{ J} \]
Step 4: Final Answer:
The change in energy is 4 J.