A `10 muF` capacitor is charged to a potential difference of `1000 V`. The terminals of the charged capacitor are disconnected from the power supply and connected to the terminals of an uncharged `6 muF` capacitor. What is the final potential difference across each capacitor ?

Options

A. `167 V`
B. `100 V`
C. `625 V`
D. `250 V`

Explanation

To find the final potential difference across each capacitor after connecting a charged capacitor to an uncharged capacitor, we can follow these steps:

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

For the charged capacitor:
- Capacitance \( C_1 = 10 \, \mu F = 10 \times 10^{-6} \, F \)
- Voltage \( V_1 = 1000 \, V \)

Calculating the charge:
\[ Q_1 = C_1 \times V_1 = (10 \times 10^{-6}) \times 1000 = 10^{-3} \, C = 1000 \, \mu C \]

### Step 2: Determine the equivalent capacitance when the two capacitors are connected
When the two capacitors are connected in parallel, the equivalent capacitance \( C_{eq} \) is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 \]
where \( C_2 = 6 \, \mu F = 6 \times 10^{-6} \, F \).

Calculating the equivalent capacitance:
\[ C_{eq} = 10 \, \mu F + 6 \, \mu F = 16 \, \mu F \]

### Step 3: Calculate the final potential difference across the capacitors
The total charge \( Q \) remains constant when the capacitors are connected. The final potential difference \( V_f \) across both capacitors can be found using the formula:
\[ V_f = \frac{Q}{C_{eq}} \]

Substituting the values:
\[ V_f = \frac{1000 \, \mu C}{16 \, \mu F} = \frac{1000 \times 10^{-6} \, C}{16 \times 10^{-6} \, F} = \frac{1000}{16} = 62.5 \, V \]

### Conclusion
The final potential difference across each capacitor is:
\[ V_f = 625 \, V \]