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. (a)0.7
B. (b)`-0.7`
C. (c)`5.6`
D. (d)`-5.6`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the initial charge on the capacitor.
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where:
- \( C \) is the capacitance,
- \( V \) is the voltage.

Given:
- \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- \( V = 20 \, V \)

Calculating the initial charge:
\[ Q = (2 \times 10^{-6} \, F) \times (20 \, V) = 40 \times 10^{-6} \, C = 40 \, \mu C \]

### Step 2: Calculate the new capacitance after the dielectric slab is removed.
When the dielectric slab is removed, the capacitance \( C' \) of the capacitor changes. The new capacitance is given by:
\[ C' = \frac{C}{K} \]
where \( K \) is the dielectric constant.

Given:
- \( K = 8 \)

Calculating the new capacitance:
\[ C' = \frac{2 \, \mu F}{8} = 0.25 \, \mu F = 0.25 \times 10^{-6} \, F \]

### Step 3: Calculate the new charge on the capacitor.
Since the battery remains connected, the voltage across the capacitor remains the same. Therefore, the new charge \( Q' \) can be calculated as:
\[ Q' = C' \times V \]

Calculating the new charge:
\[ Q' = (0.25 \times 10^{-6} \, F) \times (20 \, V) = 5 \times 10^{-6} \, C = 5 \, \mu C \]

### Step 4: Calculate the change in charge.
The change in charge \( \Delta Q \) is given by:
\[ \Delta Q = Q' - Q \]

Calculating the change in charge:
\[ \Delta Q = 5 \, \mu C - 40 \, \mu C = -35 \, \mu C \]

### Step 5: Calculate the work done by the battery.
The work done \( W \) by the battery when charge flows is given by:
\[ W = \Delta Q \times V \]

Calculating the work done:
\[ W = (-35 \times 10^{-6} \, C) \times (20 \, V) = -700 \times 10^{-6} \, J = -0.7 \, mJ \]

### Final Answer:
The work done by the battery is \( -0.7 \, mJ \).

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