Question

A capacitor of capacitance 5 \(\mu\)F is charged to 100 V and then connected to an uncharged capacitor of 2 \(\mu\)F. What is the final potential difference across the capacitors?

• A. 71.43 V
• B. 50 V
• C. 28.57 V
• D. 100 V

Hint:

When capacitors are connected in parallel after one is charged, the final potential difference can be found using charge conservation. The total charge divides inversely proportional to the capacitances.

Answer 1

The Correct Option is A

Step 1: Initialize conditions. Capacitor 1 (\( C_1 = 5 \, \mu\text{F} \)) has an initial potential difference of 100 V. Capacitor 2 (\( C_2 = 2 \, \mu\text{F} \)) is uncharged (initial charge = 0).
Step 2: Calculate initial charge on capacitor 1. Charge \( Q = C \cdot V \): \[ Q_1 = C_1 \cdot V_1 = 5 \cdot 10^{-6} \cdot 100 = 5 \cdot 10^{-4} \, \text{C}. \]
Step 3: Determine total capacitance in parallel. Capacitances add for parallel connection, resulting in a shared final potential difference: \[ C_{\text{total}} = C_1 + C_2 = 5 \, \mu\text{F} + 2 \, \mu\text{F} = 7 \, \mu\text{F}. \]
Step 4: Apply charge conservation. Initial charge \( Q_{\text{total}} = Q_1 \) redistributes between capacitors to achieve a common final potential difference \( V_f \): \[ Q_{\text{total}} = C_{\text{total}} \cdot V_f. \] Substitute values: \[ 5 \cdot 10^{-4} = 7 \cdot 10^{-6} \cdot V_f. \]
Step 5: Solve for final potential difference: \[ V_f = \frac{5 \cdot 10^{-4}}{7 \cdot 10^{-6}} = \frac{5}{7} \cdot 10^2 = \frac{500}{7} \approx 71.43 \, \text{V}. \]
Step 6: Verify result. The final potential difference (71.43 V) is less than the initial 100 V, consistent with charge sharing.