A parallel plate capacitor of `300 muF` is charged to `200 V`. If the distance between its plate is halved, what will be the potential difference between the plates and what will be the change in stored energy ?

Explanation

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

### Step 1: Calculate the initial charge on the capacitor
The formula for charge \( Q \) on a capacitor is given by:
\[ Q = C \cdot V \]
where:
- \( C = 300 \, \mu F = 300 \times 10^{-6} \, F \)
- \( V = 200 \, V \)

Substituting the values:
\[ Q = 300 \times 10^{-6} \, F \times 200 \, V = 6 \times 10^{-2} \, C \]

### Step 2: Determine the new capacitance after halving the distance
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{d} \]
When the distance \( d \) between the plates is halved, the new capacitance \( C' \) becomes:
\[ C' = \frac{\varepsilon_0 A}{d/2} = 2 \cdot \frac{\varepsilon_0 A}{d} = 2C \]
Thus,
\[ C' = 2 \times 300 \, \mu F = 600 \, \mu F \]

### Step 3: Calculate the new potential difference
The charge \( Q \) remains the same when the distance is halved. The new potential difference \( V' \) can be calculated using:
\[ V' = \frac{Q}{C'} \]
Substituting the values:
\[ V' = \frac{6 \times 10^{-2} \, C}{600 \times 10^{-6} \, F} = \frac{6 \times 10^{-2}}{600 \times 10^{-6}} = 100 \, V \]

### Step 4: Calculate the initial energy stored in the capacitor
The energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
Calculating the initial energy \( U_1 \):
\[ U_1 = \frac{1}{2} \times 300 \times 10^{-6} \, F \times (200 \, V)^2 \]
\[ U_1 = \frac{1}{2} \times 300 \times 10^{-6} \times 40000 = 6 \, J \]

### Step 5: Calculate the new energy stored in the capacitor
Now, calculate the new energy \( U_2 \) with the new capacitance and potential difference:
\[ U_2 = \frac{1}{2} C' (V')^2 \]
Substituting the values:
\[ U_2 = \frac{1}{2} \times 600 \times 10^{-6} \, F \times (100 \, V)^2 \]
\[ U_2 = \frac{1}{2} \times 600 \times 10^{-6} \times 10000 = 3 \, J \]

### Step 6: Calculate the change in stored energy
The change in stored energy \( \Delta U \) is given by:
\[ \Delta U = U_1 - U_2 \]
Substituting the values:
\[ \Delta U = 6 \, J - 3 \, J = 3 \, J \]

### Final Results
- The new potential difference between the plates is \( 100 \, V \).
- The change in stored energy is \( 3 \, J \).