Question

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm while battery is connected, it is found that it draws 25% more charge. The dielectric constant of mica is :

• A. 2
• B. 1.5
• C. 1
• D. 4

Hint:

The most important concept in such capacitor problems is to identify what remains constant.
- If the \textbf{battery remains connected}, the \textbf{voltage (V)} is constant.
- If the \textbf{battery is disconnected} before the change, the \textbf{charge (Q)} is constant.
This distinction determines the entire approach to the problem.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A capacitor is charged and remains connected to a battery. A dielectric slab is then inserted. Since the battery is still connected, the voltage (\(V\)) across the capacitor remains constant. The insertion of the dielectric increases the capacitance, causing more charge to be drawn from the battery. We are given that the charge increases by 25%.
Step 2: Key Formula or Approach:
- Charge on a capacitor: \(Q = CV\).
- Initial capacitance (air-filled): \(C_i = \frac{\epsilon_0 A}{d}\).
- Final capacitance with a dielectric slab of thickness \(t\) and dielectric constant \(K\): \(C_f = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}\).
Step 3: Detailed Explanation:
Let the initial charge be \(Q_i\) and the final charge be \(Q_f\). We are given that \(Q_f\) is 25% more than \(Q_i\).
\[ Q_f = Q_i + 0.25 Q_i = 1.25 Q_i \] Since the battery remains connected, the voltage \(V\) is constant.
\[ C_f V = 1.25 (C_i V) \] \[ C_f = 1.25 C_i = \frac{5}{4} C_i \] Now, let's express the capacitances in terms of the physical dimensions.
- Plate separation \(d = 5\) mm.
- Thickness of mica sheet \(t = 2\) mm.
Initial capacitance:
\[ C_i = \frac{\epsilon_0 A}{5} \] Final capacitance (the remaining air gap is \(d-t = 3\) mm):
\[ C_f = \frac{\epsilon_0 A}{(d-t) + \frac{t}{K}} = \frac{\epsilon_0 A}{3 + \frac{2}{K}} \] Now, substitute these into the relation \(C_f = 1.25 C_i\):
\[ \frac{\epsilon_0 A}{3 + \frac{2}{K}} = 1.25 \times \left( \frac{\epsilon_0 A}{5} \right) \] Cancel out the term \(\epsilon_0 A\) from both sides:
\[ \frac{1}{3 + \frac{2}{K}} = \frac{1.25}{5} = \frac{1}{4} \] Invert both sides:
\[ 3 + \frac{2}{K} = 4 \] \[ \frac{2}{K} = 1 \] \[ K = 2 \] Step 4: Final Answer:
The dielectric constant of the mica sheet is 2.