Question

For sufficient time the key \( S_1 \) is closed and \( S_2 \) is open. Now key \( S_2 \) is closed and \( S_1 \) is open. What is the final charge on the capacitor?

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

Hint:

The charge on a capacitor does not change unless there is a path for the current to flow, such as a discharge path or an external resistor.

Answer 1

The Correct Option is B

To determine the final charge on the capacitor, we need to consider the given scenario involving two switches, \( S_1 \) and \( S_2 \). Initially, \( S_1 \) is closed, allowing the capacitor to charge, while \( S_2 \) is open. When \( S_1 \) is opened and \( S_2 \) is closed, the charge on the capacitor can be evaluated using the concept of charge conservation in an RC circuit.

Given:

• \( Q = C \times V \)

Where:

• \( Q \) is the charge on the capacitor
• \( C \) is the capacitance
• \( V \) is the voltage across the capacitor

In the initial condition, with \( S_1 \) closed, the capacitor charges to a maximum charge \( Q = C \times V \). Now, without loss of generality, suppose:

• \( V = 5 \, \text{V} \) (voltage across the capacitor when fully charged)
• \( C = 1 \, \text{mF} \) (capacitance of the capacitor)

Thereafter, when \( S_2 \) is closed and \( S_1 \) is open, the charge on the capacitor equals the initial charge \( CV \). Therefore:

• \( Q = 1 \, \text{mF} \times 5 \, \text{V} = 5 \times 10^{-3} \, \text{C} = 5 \, \text{mC} \)

Thus, the final charge on the capacitor, after \( S_2 \) is closed and \( S_1 \) is open, is:

• 5 mC