A parallel plate capacitor has a capacitance of `2 mu F`. A slab of dielectric constant 5 is inserted between the plates and the capacitor is charged to `100 V` and then isolated . (a) What is the new potential diff., if the dielectric slab is removed ? (b) How much work is required to remove teh dielectric slab ?

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as per the question.

### Given Data:
- Capacitance of the capacitor, \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Dielectric constant, \( K = 5 \)
- Initial voltage, \( V_1 = 100 \, V \)

### Part (a): New Potential Difference after Removing the Dielectric Slab

1. **Calculate the charge on the capacitor when the dielectric is present**:
\[ Q = C \times V_1 = (2 \times 10^{-6} \, F) \times (100 \, V) = 2 \times 10^{-4} \, C = 200 \, \mu C \]

2. **Calculate the new capacitance without the dielectric**:
The capacitance without the dielectric is given by:
\[ C_2 = \frac{C_1}{K} = \frac{C}{K} = \frac{2 \times 10^{-6} \, F}{5} = 0.4 \times 10^{-6} \, F = 0.4 \, \mu F \]

3. **Calculate the new potential difference after removing the dielectric**:
The potential difference when the dielectric is removed can be calculated using the formula:
\[ V' = \frac{Q}{C_2} = \frac{200 \times 10^{-6} \, C}{0.4 \times 10^{-6} \, F} = 500 \, V \]

### Part (b): Work Required to Remove the Dielectric Slab

1. **Calculate the initial energy stored in the capacitor with the dielectric**:
\[ U_1 = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times (2 \times 10^{-6} \, F) \times (100 \, V)^2 = \frac{1}{2} \times 2 \times 10^{-6} \times 10000 = 0.01 \, J \]

2. **Calculate the energy stored in the capacitor without the dielectric**:
\[ U_2 = \frac{1}{2} C_2 V'^2 = \frac{1}{2} \times (0.4 \times 10^{-6} \, F) \times (500 \, V)^2 = \frac{1}{2} \times 0.4 \times 10^{-6} \times 250000 = 0.05 \, J \]

3. **Calculate the work done to remove the dielectric**:
The work done is the change in energy:
\[ W = U_2 - U_1 = 0.05 \, J - 0.01 \, J = 0.04 \, J \]

### Final Answers:
(a) The new potential difference after removing the dielectric slab is **500 V**.
(b) The work required to remove the dielectric slab is **0.04 J**.