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 find the value of \( X \) in the electric field \( E = 10^X \, \text{N/C} \) for a parallel-plate capacitor with given parameters, we can follow these steps:

### Step 1: Write down the given values
- Capacitance \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Energy stored \( U = 10 \, mJ = 10 \times 10^{-3} \, J \)
- Separation between the plates \( d = 1 \, mm = 1 \times 10^{-3} \, m \)

### Step 2: Use the formula for energy stored in a capacitor
The energy stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
From this, we can solve for the potential difference \( V \):
\[ V^2 = \frac{2U}{C} \]
\[ V = \sqrt{\frac{2U}{C}} \]

### Step 3: Substitute the values into the equation
Substituting the values of \( U \) and \( C \):
\[ V = \sqrt{\frac{2 \times (10 \times 10^{-3})}{2 \times 10^{-6}}} \]
\[ V = \sqrt{\frac{20 \times 10^{-3}}{2 \times 10^{-6}}} \]
\[ V = \sqrt{10^3} = \sqrt{1000} = 31.62 \, V \quad (\text{approximately}) \]

### Step 4: Calculate the electric field \( E \)
The electric field \( E \) between the plates of the capacitor is given by:
\[ E = \frac{V}{d} \]
Substituting the values of \( V \) and \( d \):
\[ E = \frac{31.62}{1 \times 10^{-3}} = 31620 \, N/C \]

### Step 5: Express \( E \) in the form \( 10^X \)
We can express \( 31620 \, N/C \) in scientific notation:
\[ E = 3.162 \times 10^4 \, N/C \]
Thus, we can write:
\[ E = 10^{4.5} \, N/C \]
From this, we can see that \( X \) is approximately \( 5 \).

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

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