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: Voltage divider rule:
For capacitors in series, $V_1 : V_2 = \frac{1}{C_1} : \frac{1}{C_2} = C_2 : C_1$.

Step 2: Share:
So $V_2 = V\times\frac{C_1}{C_1 + C_2}$ (B).

Final Answer:
$\frac{C_1V}{C_1 + C_2}$. \[ \boxed{\frac{C_1V}{C_1 + C_2}} \]