A capacitor of capacitance `2muF` has a dielectric slab of dielectric constant 8 between its plates. The slab covers the entire volume between the plates. This capacitor is connected to a battery if EMF 20 V and fully charged Now with the battery still connected, the slab is removed from the capactor. during the removal, the work done by the battery is (in mJ).

Options

A. 0.7
B. `-0.7`
C. `5.6`
D. `-5.6`

Explanation

To solve the problem, we need to find the work done by the battery when the dielectric slab is removed from the capacitor while it is still connected to the battery.

### Step-by-Step Solution:

1. **Calculate the initial capacitance with the dielectric**:
The capacitance \( C \) of a capacitor with a dielectric is given by:
\[ C = K \cdot C_0 \]
where \( K \) is the dielectric constant and \( C_0 \) is the capacitance without the dielectric. Here, \( C_0 = 2 \mu F \) and \( K = 8 \).
\[ C = 8 \cdot 2 \mu F = 16 \mu F \]

2. **Calculate the initial charge on the capacitor**:
The charge \( Q \) on a capacitor is given by:
\[ Q = C \cdot V \]
where \( V \) is the voltage across the capacitor. Here, \( V = 20 V \).
\[ Q = 16 \mu F \cdot 20 V = 320 \mu C \]

3. **Calculate the final capacitance without the dielectric**:
When the dielectric slab is removed, the capacitance returns to its original value:
\[ C' = C_0 = 2 \mu F \]

4. **Calculate the final charge on the capacitor**:
The charge on the capacitor after the dielectric is removed can be calculated using the same formula:
\[ Q' = C' \cdot V = 2 \mu F \cdot 20 V = 40 \mu C \]

5. **Calculate the change in charge**:
The change in charge \( \Delta Q \) when the dielectric slab is removed is:
\[ \Delta Q = Q - Q' = 320 \mu C - 40 \mu C = 280 \mu C \]

6. **Calculate the work done by the battery**:
The work done \( W \) by the battery when the dielectric is removed can be calculated using the formula:
\[ W = V \cdot \Delta Q \]
Substituting the values:
\[ W = 20 V \cdot 280 \mu C = 20 \cdot 280 \times 10^{-6} J = 0.0056 J \]
Converting to millijoules:
\[ W = 0.0056 J \times 1000 = 5.6 mJ \]

### Final Answer:
The work done by the battery during the removal of the slab is **5.6 mJ**.