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: Identify the initial conditions. The first capacitor (\( C_1 = 5 \, \mu\text{F} \)) is charged to a potential difference of 100 V. The second capacitor (\( C_2 = 2 \, \mu\text{F} \)) is uncharged, so its initial charge is 0.
Step 2: Calculate the initial charge on the first capacitor. The charge \( Q \) on a capacitor is given by \( Q = C \cdot V \): \[ Q_1 = C_1 \cdot V_1 = 5 \cdot 10^{-6} \cdot 100 = 5 \cdot 10^{-4} \, \text{C}. \]
Step 3: Determine the total capacitance when connected in parallel. Since the capacitors are connected together, they share the same final potential difference, and their capacitances add up: \[ C_{\text{total}} = C_1 + C_2 = 5 \, \mu\text{F} + 2 \, \mu\text{F} = 7 \, \mu\text{F}. \]
Step 4: Apply the conservation of charge. When connected, the charge from the first capacitor redistributes between both capacitors. The total initial charge \( Q_{\text{total}} = Q_1 \) is conserved: \[ Q_{\text{total}} = C_{\text{total}} \cdot V_f, \] where \( V_f \) is the final potential difference. Substitute the values: \[ 5 \cdot 10^{-4} = 7 \cdot 10^{-6} \cdot V_f. \]
Step 5: Solve for the 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 the result. The final potential difference should be less than the initial 100 V due to charge sharing, and the calculated value 71.43 V is consistent with the options.