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 the principles of capacitors and charge conservation.

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

Using the formula for charge \( Q = C \times V \):
\[ Q_1 = C_1 \times V_1 = 2 \, \mu F \times 200 \, V = 400 \, \mu C \]

### Step 2: Understand the configuration after combining capacitors
After disconnecting the battery, the charged capacitor is combined in parallel with an uncharged capacitor. The total charge remains the same, and the potential difference across both capacitors is now \( V_f = 40 \, V \).

### Step 3: Write the equation for charge conservation
The total charge before combining the capacitors must equal the total charge after combining them.

Let \( C_2 \) be the capacitance of the second capacitor (the uncharged one). The charge on the first capacitor remains \( Q_1 = 400 \, \mu C \), and the charge on the second capacitor when the voltage across it is \( 40 \, V \) is given by:
\[ Q_2 = C_2 \times V_f = C_2 \times 40 \, V \]

By conservation of charge:
\[ Q_1 = Q_2 + Q_3 \]
Where \( Q_3 \) is the charge on the first capacitor after they are connected:
\[ 400 \, \mu C = 2 \, \mu F \times 40 \, V + C_2 \times 40 \, V \]

### Step 4: Substitute and simplify
Calculate the charge on the first capacitor after combining:
\[ Q_3 = 2 \, \mu F \times 40 \, V = 80 \, \mu C \]

Now substitute this into the charge conservation equation:
\[ 400 \, \mu C = 80 \, \mu C + C_2 \times 40 \, V \]

### Step 5: Solve for \( C_2 \)
Rearranging the equation gives:
\[ 400 \, \mu C - 80 \, \mu C = C_2 \times 40 \, V \]
\[ 320 \, \mu C = C_2 \times 40 \, V \]
\[ C_2 = \frac{320 \, \mu C}{40 \, V} = 8 \, \mu F \]

### Final Answer
The capacitance of the second capacitor is \( C_2 = 8 \, \mu F \).

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