Question

A parallel plate capacitor has 1 \(\mu\)F capacitance. One of its two plates is given \(+2 \mu C\) charge and the other plate, \(+4 \mu C\) charge. The potential difference developed across the capacitor is:

• A. 3 V
• B. 1 V
• C. 5 V
• D. 2 V

Hint:

- The charge on the inner plates of a capacitor is given by \( \frac{|Q_1 - Q_2|}{2} \).
- The potential difference across a capacitor is calculated using \(V = \frac{Q}{C}\).
- Always ensure proper sign conventions when dealing with capacitors in circuits.

Answer 1

The Correct Option is B

Step 1: Understanding the Capacitance Formula
The relationship between charge \(Q\), capacitance \(C\), and potential difference \(V\) is given by the formula: \[ V = \frac{Q}{C} \] Where:
- \(Q\) is the charge on the plates.

- \(C\) is the capacitance of the parallel plate capacitor.

- \(V\) is the potential difference across the plates.

Step 2: Charge on the Inner Plates
According to Gauss’s law, the charge appearing on the inner plates of the capacitor is given by: \[ Q_{\text{inner}} = \frac{|Q_1 - Q_2|}{2} \]
Given \(Q_1 = 4 \mu C\) and \(Q_2 = 2 \mu C\), we get: \[ Q_{\text{inner}} = \frac{|4 - 2|}{2} = \frac{2}{2} = 1 \mu C \]
Step 3: Calculating the Potential Difference
The given capacitance \(C = 1 \mu F\). Now, using the formula: \[ V = \frac{Q_{\text{inner}}}{C} \] \[ V = \frac{1 \mu C}{1 \mu F} = 1 V \] Final Answer: The potential difference developed across the capacitor is 1 V.