A capacitor of capacitance of `2muF` is charged to a potential difference of `200V`, after disconnecting from the battery, it is connected in parallel with another uncharged capacitor. The final common potential is `20V` then the capacitance of second capacitor is :

Options

A. `2muF`
B. `4muF`
C. `18muF`
D. `16muF`

Answer

To solve the problem step by step, we will follow the principles of capacitance, charge, and potential difference.

### Step 1: Calculate the charge on the first capacitor
The formula for charge \( Q \) on a capacitor is given by:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the potential difference.

Given:
- Capacitance \( C_1 = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Voltage \( V_1 = 200 \, V \)

Calculating the charge:
\[ Q = 2 \times 10^{-6} \, F \times 200 \, V = 4 \times 10^{-4} \, C = 400 \, \mu C \]

### Step 2: Understand the situation after disconnecting the battery
After charging, the capacitor is disconnected from the battery and then connected in parallel to an uncharged capacitor \( C_2 \). The total charge remains the same, but it will now distribute between the two capacitors.

### Step 3: Use the final common potential
When the two capacitors are connected in parallel, they will have the same final potential \( V_f \). Given:
\[ V_f = 20 \, V \]

### Step 4: Write the equation for total charge
The total charge \( Q \) is conserved, so we can express it as:
\[ Q = Q_1 + Q_2 \]
where \( Q_1 \) is the charge on the first capacitor and \( Q_2 \) is the charge on the second capacitor.

The charge on each capacitor can be expressed as:
\[ Q_1 = C_1 \times V_f = 2 \, \mu F \times 20 \, V = 40 \, \mu C \]
\[ Q_2 = C_2 \times V_f \]

### Step 5: Set up the equation
Since the total charge remains constant:
\[ 400 \, \mu C = 40 \, \mu C + Q_2 \]
Substituting for \( Q_2 \):
\[ Q_2 = 400 \, \mu C - 40 \, \mu C = 360 \, \mu C \]

### Step 6: Relate charge and capacitance for the second capacitor
Using the charge on the second capacitor:
\[ Q_2 = C_2 \times V_f \]
Substituting the known values:
\[ 360 \, \mu C = C_2 \times 20 \, V \]

### Step 7: Solve for \( C_2 \)
Rearranging the equation gives:
\[ C_2 = \frac{360 \, \mu C}{20 \, V} = 18 \, \mu F \]

### Final Answer
The capacitance of the second capacitor \( C_2 \) is:
\[ \boxed{18 \, \mu F} \]

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