A condenser of capacitance `10 muF` has been charged to `100V`. It is now connected to another uncharged condenser in parallel. The common potential becomes `40 V`. The capacitance of another condenser is

Options

A. (a)`15 muF`
B. (b)`5 muF`
C. (c)`10 muF`
D. (d)`16 umF`

Answer

To solve the problem step by step, we will use the concepts of capacitance, charge, and the behavior of capacitors in parallel.

### Step 1: Calculate the charge on the first capacitor.
The first capacitor has a capacitance \( C_1 = 10 \, \mu F \) and is charged to a voltage \( V_1 = 100 \, V \).

The charge \( Q_1 \) on the capacitor can be calculated using the formula:
\[ Q = C \times V \]
Substituting the values:
\[ Q_1 = 10 \, \mu F \times 100 \, V = 1000 \, \mu C \]

### Step 2: Understand the setup after connecting the second capacitor.
When the charged capacitor is connected in parallel to an uncharged capacitor, they will share the charge until they reach a common voltage \( V_f \). In this case, the common voltage is given as \( V_f = 40 \, V \).

### Step 3: Write the charge conservation equation.
The total charge before connecting the capacitors must equal the total charge after connecting them. Thus, we have:
\[ Q_{total} = Q_1 + Q_2 \]
Where \( Q_2 \) is the charge on the second capacitor after they are connected. The charge on the second capacitor can be expressed as:
\[ Q_2 = C_2 \times V_f \]
Where \( C_2 \) is the capacitance of the second capacitor, which we need to find.

### Step 4: Set up the equation using the known values.
Since the total charge is conserved, we can write:
\[ Q_1 = C_1 \times V_1 = 1000 \, \mu C \]
And after connecting:
\[ Q_{total} = Q_1 = Q_1 + Q_2 = 1000 \, \mu C \]
Substituting \( Q_2 \):
\[ 1000 \, \mu C = 1000 \, \mu C + C_2 \times 40 \, V \]

### Step 5: Solve for \( C_2 \).
Rearranging the equation gives:
\[ C_2 \times 40 = 1000 - 400 \]
\[ C_2 \times 40 = 600 \]
\[ C_2 = \frac{600}{40} = 15 \, \mu F \]

### Conclusion
The capacitance of the second capacitor is \( C_2 = 15 \, \mu F \).