The capacity of a parallel plate condenser with air as dielectric is `2muF` . The space between the plates is filled with dielectric slab with K = 5. It is charged to a potential of 200V and disconnected from cell. Work done in removing the slab from the condenser completely

Options

A. 0.8 J
B. 0.6 J
C. 1.2J
D. 1.6J

Answer

To solve the problem step by step, we will calculate the work done in removing the dielectric slab from a parallel plate capacitor.

### Step 1: Calculate the initial energy stored in the capacitor (U1)
The initial capacitance \( C_1 \) is given as \( 2 \, \mu F = 2 \times 10^{-6} \, F \) and the initial voltage \( V_1 \) is \( 200 \, V \).

The energy stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]

Substituting the values:
\[ U_1 = \frac{1}{2} \times (2 \times 10^{-6}) \times (200)^2 \]
Calculating \( (200)^2 = 40000 \):
\[ U_1 = \frac{1}{2} \times (2 \times 10^{-6}) \times 40000 \]
\[ U_1 = (1 \times 10^{-6}) \times 40000 = 0.04 \, J = 4 \times 10^{-2} \, J \]

### Step 2: Calculate the final capacitance (C2) after removing the dielectric slab
When the dielectric slab with dielectric constant \( K = 5 \) is removed, the capacitance changes. The new capacitance \( C_2 \) can be calculated as:
\[ C_2 = \frac{C_1}{K} = \frac{2 \times 10^{-6}}{5} = 0.4 \times 10^{-6} \, F = 0.4 \, \mu F \]

### Step 3: Calculate the new voltage (V2) after removing the dielectric slab
Since the capacitor is disconnected from the cell, the charge \( Q \) remains constant. The initial charge \( Q \) can be calculated as:
\[ Q = C_1 V_1 = (2 \times 10^{-6}) \times 200 = 4 \times 10^{-4} \, C \]

Now, using the new capacitance \( C_2 \), we can find the new voltage \( V_2 \):
\[ V_2 = \frac{Q}{C_2} = \frac{4 \times 10^{-4}}{0.4 \times 10^{-6}} = 1000 \, V \]

### Step 4: Calculate the final energy stored in the capacitor (U2)
Using the new capacitance \( C_2 \) and the new voltage \( V_2 \):
\[ U_2 = \frac{1}{2} C_2 V_2^2 \]
Substituting the values:
\[ U_2 = \frac{1}{2} \times (0.4 \times 10^{-6}) \times (1000)^2 \]
Calculating \( (1000)^2 = 1000000 \):
\[ U_2 = \frac{1}{2} \times (0.4 \times 10^{-6}) \times 1000000 \]
\[ U_2 = (0.2 \times 10^{-6}) \times 1000000 = 0.2 \, J \]

### Step 5: Calculate the work done in removing the slab
The work done \( W \) in removing the dielectric slab is the change in energy:
\[ W = U_2 - U_1 \]
Substituting the values:
\[ W = 0.2 - 0.04 = 0.16 \, J \]

### Final Answer
The work done in removing the slab from the condenser completely is \( 0.16 \, J \).

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