Question

Two indentical metal plates are given charges \(q_1\) and \(q_2\) \((q_2 < q_1)\) respectively. They are brought close together to form a parallel plate capacitor with capacitance 'C'. The potential difference 'V' between the plates is

• A. \(\frac{q_1-q_2}{C}\)
• B. \(\frac{q_1+q_2}{C}\)
• C. \(\frac{q_1-q_2}{2C}\)
• D. \(\frac{q_1+q_2}{2C}\)

Hint:

In series the charge is the same, so voltage divides inversely with capacitance.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a series combination, the same charge Q is on each capacitor. The equivalent capacitance is \(C_s = \frac{C_1C_2}{C_1 + C_2}\).

Step 2: Charge:
\(Q = C_sV = \frac{C_1C_2V}{C_1 + C_2}\).

Step 3: Voltage across \(C_2\):
\[ V_2 = \frac{Q}{C_2} = \frac{C_1V}{C_1 + C_2} \]
The smaller capacitor takes the larger share of the voltage, and the answer has \(C_1\) in the numerator for \(V_2\). Option (D), \(\frac{C_2V}{C_1+C_2}\), is the voltage across \(C_1\).

Final Answer:
The voltage across \(C_2\) is \(\frac{C_1V}{C_1 + C_2}\), option (B). \[ \boxed{\frac{C_1V}{C_1 + C_2}} \]