1. Given data

• Capacitance of the parallel plate capacitor: $C = 1\,\mu F$
• Charge on one plate: $q_1 = +2\,\mu C$
• Charge on the other plate: $q_2 = +4\,\mu C$

2. Key concept

For a capacitor, the potential difference depends on the charges on the two plates. If the plates carry charges that are not equal and opposite, we can write the charges as:

• Common charge contributing to capacitor action: $\pm q$
• Excess charge distributed equally as a net charge on the capacitor system.

Thus, $q = \frac{|q_2-q_1|}{2}$

Here, $q = \frac{4-2}{2} = 1\,\mu C$

This is the effective charge responsible for the electric field between the plates.

3. Potential difference across capacitor

Using $V = \frac{q}{C}$

we get $V = \frac{1\,\mu C}{1\,\mu F} = 1\,V$

4. Checking options

• A: $1\,V$ ✅
• B: $5\,V$ ❌
• C: $2\,V$ ❌
• D: $3\,V$ ❌

5. Final answer

The potential difference developed across the capacitor is: $\boxed{1\,V}$