To find the potential difference developed across a parallel plate capacitor, we can use the relationship between charge (Q), capacitance (C), and potential difference (V). The formula is given by:
\[ V = \frac{Q}{C} \]
### Step-by-Step Solution:
1. **Identify the Given Values:**
- Capacitance, \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)
- Charge on one plate, \( Q_1 = +2 \, \mu C = 2 \times 10^{-6} \, C \)
- Charge on the other plate, \( Q_2 = +4 \, \mu C = 4 \times 10^{-6} \, C \)
2. **Determine the Net Charge on the Capacitor:**
- The net charge \( Q \) on the capacitor is the difference between the charges on the two plates:
\[
Q = Q_2 - Q_1 = 4 \, \mu C - 2 \, \mu C = 2 \, \mu C = 2 \times 10^{-6} \, C
\]
3. **Calculate the Potential Difference:**
- Now, we can use the formula for potential difference:
\[
V = \frac{Q}{C}
\]
- Substitute the values:
\[
V = \frac{2 \times 10^{-6} \, C}{1 \times 10^{-6} \, F} = 2 \, V
\]
4. **Conclusion:**
- The potential difference developed across the capacitor is \( 2 \, V \).
### Final Answer:
The potential difference developed across the capacitor is **2 volts**.
