A capacitor of 10 `muF` is connected to a source of 50 V and then disconnected. When the space between the plates is filled with teflon of dielectric constant 2.5. The new potential difference is -

Options

A. 20 V
B. 20.8 V
C. 23.8 V
D. 25 V

Explanation

To solve the problem step by step, we will follow the principles of capacitors and dielectrics.

### Step 1: Understand the initial conditions
We have a capacitor with a capacitance \( C = 10 \, \mu F \) connected to a voltage source of \( V = 50 \, V \). When connected, the charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C \cdot V \]

### Step 2: Calculate the initial charge
Substituting the values into the formula, we get:
\[ Q = 10 \times 10^{-6} \, F \cdot 50 \, V = 5 \times 10^{-4} \, C \]
So, the initial charge \( Q \) on the capacitor is \( 5 \times 10^{-4} \, C \).

### Step 3: Disconnect the battery
After disconnecting the battery, the charge \( Q \) on the capacitor remains constant because there is no external circuit to change it.

### Step 4: Introduce the dielectric
Now, we fill the space between the capacitor plates with a dielectric material (Teflon) that has a dielectric constant \( K = 2.5 \). The new capacitance \( C' \) of the capacitor with the dielectric is given by:
\[ C' = K \cdot C = 2.5 \cdot 10 \, \mu F = 25 \, \mu F \]

### Step 5: Relate charge, capacitance, and voltage
The relationship between charge, capacitance, and voltage is given by:
\[ Q = C' \cdot V' \]
Since \( Q \) remains constant, we can set up the equation:
\[ 5 \times 10^{-4} \, C = 25 \times 10^{-6} \, F \cdot V' \]

### Step 6: Solve for the new voltage \( V' \)
Rearranging the equation to solve for \( V' \):
\[ V' = \frac{Q}{C'} = \frac{5 \times 10^{-4} \, C}{25 \times 10^{-6} \, F} \]
Calculating \( V' \):
\[ V' = \frac{5 \times 10^{-4}}{25 \times 10^{-6}} = \frac{5}{25} \times 10^{2} = 0.2 \times 100 = 20 \, V \]

### Conclusion
The new potential difference \( V' \) after filling the capacitor with Teflon is \( 20 \, V \).

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