Question

The final charge on the capacitor, when key \( S_1 \) is closed and \( S_2 \) is open, is:

• A. 5 \( \mu C \)
• B. 5 mC
• C. 25 mC
• D. 0.1 C

Hint:

The final charge on a capacitor is the product of its capacitance and the voltage across it.

Answer 1

The Correct Option is B

To determine the final charge on the capacitor when key \( S_1 \) is closed and \( S_2 \) is open, we need to consider the circuit configuration and basic principles of capacitance. When \( S_1 \) is closed, the capacitor is connected directly across the voltage source, and \( S_2 \) is open, meaning it is not part of the circuit. The charge \( Q \) on a capacitor is given by the equation:

\(Q = C \cdot V\)

Where Q is the charge, C is the capacitance of the capacitor, and V is the voltage across the capacitor. According to the provided options and the solution, if the correct answer is 5 mC (milllicoulombs), we can infer:
\( Q = 5 \, \text{mC} = 5 \times 10^{-3} \, \text{C} \)

To achieve a final charge of 5 mC, the combination of values for capacitance \( C \) and voltage \( V \) in the equation is such that their product equals \( 5 \times 10^{-3} \) C. Therefore, the final charge, when \( S_1 \) is closed and \( S_2 \) is open, in this specific scenario, is 5 mC.