A `2 muF` condenser is charged upto 200 V and then battery is removed. On combining this with another uncharged condenser in parallel, the potential differences between two plates are found to be 40 V. The capacity of second condenser is

Options

A. `2 muF`
B. `4 mu F`
C. `8 mu F`
D. `16 mu F`

Answer

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

### Step 1: Calculate the initial charge on the first capacitor
The first capacitor has a capacitance \( C_1 = 2 \mu F \) and is charged to a voltage \( V_1 = 200 V \). The charge \( Q \) on a capacitor is given by the formula:

\[ Q = C \times V \]

Substituting the values:

\[ Q_1 = 2 \mu F \times 200 V = 2 \times 10^{-6} F \times 200 V = 400 \mu C \]

### Step 2: Understand the configuration after combining the capacitors
After the battery is removed, the charged capacitor is connected in parallel with an uncharged capacitor \( C_2 \). The final potential difference across both capacitors is given as \( V_f = 40 V \).

### Step 3: Apply the conservation of charge
The total charge before combining the capacitors must equal the total charge after they are combined. Initially, we have:

\[ Q_{\text{initial}} = Q_1 = 400 \mu C \]

After combining, the charge on the first capacitor is:

\[ Q_{1f} = C_1 \times V_f = 2 \mu F \times 40 V = 80 \mu C \]

Let the charge on the second capacitor be \( Q_{2f} \). By conservation of charge:

\[ Q_{\text{initial}} = Q_{1f} + Q_{2f} \]

Substituting the values:

\[ 400 \mu C = 80 \mu C + Q_{2f} \]

### Step 4: Solve for the charge on the second capacitor
Rearranging the equation gives:

\[ Q_{2f} = 400 \mu C - 80 \mu C = 320 \mu C \]

### Step 5: Calculate the capacitance of the second capacitor
The charge on the second capacitor is related to its capacitance and the final voltage by the formula:

\[ Q_{2f} = C_2 \times V_f \]

Substituting the known values:

\[ 320 \mu C = C_2 \times 40 V \]

To find \( C_2 \):

\[ C_2 = \frac{320 \mu C}{40 V} = 8 \mu F \]

### Final Answer
The capacitance of the second capacitor is:

\[ C_2 = 8 \mu F \]

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