1. Let the final charges on the facing plates be found using symmetry

When two identical metal plates are brought close to form a capacitor, charge redistributes.

Let the charges appearing on the outer surfaces be $x$ and $y$, and the charges on the inner facing surfaces be $+Q$ and $-Q$ respectively.

Since the plates are identical and placed symmetrically, the outer surfaces must have equal charge density, so $x=y.$

Now apply charge conservation to each plate:

• Left plate total charge: $x+Q=q_1$
• Right plate total charge: $y-Q=q_2$

Since $x=y$, subtract the second equation from the first: $(x+Q)-(x-Q)=q_1-q_2$ $2Q=q_1-q_2$ $Q=\frac{q_1-q_2}{2}$

So the capacitor effectively has charge magnitude $Q=\frac{q_1-q_2}{2}.$

2. Use the capacitor relation

For a capacitor, $V=\frac{Q}{C}$ where $Q$ is the magnitude of charge on either plate of the capacitor.

Thus, $V=\frac{1}{C}\cdot \frac{q_1-q_2}{2}$ $V=\frac{q_1-q_2}{2C}$

3. Check the options

The obtained expression is $\frac{q_1-q_2}{2C}$ which matches Option C.

4. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

They agree.