A charge of 20 microCoulomb is placed on the positive plate of on isolated parallel - plate capacitor of capacitance `10 microFarad` calculate the potential difference developed between the plates .

Explanation

To solve the problem of finding the potential difference developed between the plates of an isolated parallel-plate capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Charge on the positive plate, \( Q = 20 \, \mu C = 20 \times 10^{-6} \, C \)
- Capacitance of the capacitor, \( C = 10 \, \mu F = 10 \times 10^{-6} \, F \)

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