The electrostatic energy stored in a parallel-plate capacitor of capacitance `2muF` is 10mJ. If the separation between the plates of the capacitor is 1mm, the elctric field inside the capacitor is `10^(X)N//C`. The value of X is_____.

Explanation

To solve the problem, we need to find the electric field inside a parallel-plate capacitor given its capacitance, stored energy, and the separation between the plates.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Capacitance (C) = 2 µF = \(2 \times 10^{-6}\) F
- Energy (U) = 10 mJ = \(10 \times 10^{-3}\) J
- Separation between plates (d) = 1 mm = \(1 \times 10^{-3}\) m

2. **Use the Formula for Energy Stored in a Capacitor**:
The energy stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
Rearranging this formula to solve for the potential difference (V):
\[ V = \sqrt{\frac{2U}{C}} \]

3. **Substitute the Known Values into the Formula**:
\[ V = \sqrt{\frac{2 \times (10 \times 10^{-3})}{2 \times 10^{-6}}} \]
Simplifying this:
\[ V = \sqrt{\frac{20 \times 10^{-3}}{2 \times 10^{-6}}} = \sqrt{10^3} = \sqrt{1000} = 31.62 \text{ V} \]

4. **Calculate the Electric Field (E)**:
The electric field (E) in a parallel-plate capacitor is given by:
\[ E = \frac{V}{d} \]
Substituting the values we have:
\[ E = \frac{31.62}{1 \times 10^{-3}} = 31620 \text{ N/C} \]

5. **Express the Electric Field in the Form \(10^X\)**:
We can express \(31620\) as:
\[ E = 3.162 \times 10^4 \text{ N/C} \]
Thus, we can write:
\[ E = 10^{5} \text{ N/C} \]
Therefore, \(X = 5\).

### Final Answer:
The value of \(X\) is **5**.