Question

Two identical metal plates are given charges $q_1$ and $q_2$ ($q_2 < q_1$) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance 'C', the potential difference 'V' between the plates is \dots

• 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:

A capacitor's voltage is solely driven by the difference in charge between its plates. The common "average" charge $\frac{q_1+q_2}{2}$ gets pushed to the outer surfaces and contributes absolutely nothing to the potential difference across the gap!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Two large, parallel conducting plates are given arbitrary initial charges and brought close to form a capacitor. We must find the resulting potential difference between them.

Step 2: Key Formula or Approach:
When two large parallel plates are placed facing each other, the charges redistribute themselves on the inner and outer surfaces to ensure the electric field inside the bulk of the metal is zero.
A well-known derived result is:
- The charge on the outer-facing surfaces of both plates will be equal: $Q_{outer} = \frac{q_1 + q_2}{2}$.
- The charge on the inner-facing surfaces will be equal and opposite, forming the effective capacitor charge: $q_{inner} = \pm \frac{q_1 - q_2}{2}$.
The potential difference is solely determined by this inner charge configuration: $V = \frac{Q_{effective}}{C}$.

Step 3: Detailed Explanation:
Let's find the effective charge ($Q_{eff}$) residing on the inner surfaces of the capacitor plates.
Plate 1's inner charge = $\frac{q_1 - q_2}{2}$
Plate 2's inner charge = $\frac{q_2 - q_1}{2} = - \left( \frac{q_1 - q_2}{2} \right)$
This equal and opposite charge distribution creates the uniform electric field between the plates.
Thus, the actual "stored charge" of this capacitor is $Q_{eff} = \frac{q_1 - q_2}{2}$.
Using the fundamental capacitor equation $Q = CV$:
$$V = \frac{Q_{eff}}{C} = \frac{\frac{q_1 - q_2}{2}}{C} = \frac{q_1 - q_2}{2C}$$

Step 4: Final Answer:
The potential difference is $\frac{q_1 - q_2}{2C}$, matching option (c).