A parallel plate condenser is connected to a battery of e.m.f. 4 volt. If a plate of dielectric constant inserted into it, then the potential difference on condenser will be

Options

A. 0.5 V
B. 2 V
C. 4V
D. 32 V

Answer

To solve the problem, we need to understand how the insertion of a dielectric affects the potential difference across a parallel plate capacitor connected to a battery.

### Step-by-Step Solution:

1. **Understanding the Initial Setup:**
- We have a parallel plate capacitor connected to a battery with an electromotive force (e.m.f.) of 4 volts.
- The initial capacitance of the capacitor is denoted as \( C \).

2. **Initial Charge on the Capacitor:**
- The charge \( Q \) on the capacitor can be expressed using the formula:
\[ Q = C \times E \]
where \( E \) is the e.m.f. of the battery. Here, \( E = 4 \) volts.
- Therefore, the initial charge on the capacitor is:
\[ Q = C \times 4 \]

3. **Inserting the Dielectric:**
- When a dielectric material with dielectric constant \( K \) is inserted into the capacitor, the capacitance increases. The new capacitance \( C' \) is given by:
\[ C' = K \times C \]

4. **Effect on Charge with Dielectric:**
- The charge on the capacitor after inserting the dielectric can be calculated as:
\[ Q' = C' \times E = (K \times C) \times 4 = 4K \times C \]

5. **Potential Difference Across the Capacitor:**
- Despite the change in capacitance and charge, the potential difference \( V \) across the capacitor remains equal to the e.m.f. of the battery because the capacitor is still connected to the battery. Therefore:
\[ V = E = 4 \text{ volts} \]

6. **Conclusion:**
- The potential difference across the capacitor after inserting the dielectric remains unchanged at 4 volts.

### Final Answer:
The potential difference on the condenser will be **4 volts**.

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