separation between the plates of a parallel-plate capacitor is 2 mm and the area of its plates is `5cm^(2)`. If the capacitor is charged such that it has 0.01J energy stored in it, the electrostatic force of attraction between its plates is _______N.

Explanation

To solve the problem, we need to find the electrostatic force of attraction between the plates of a parallel-plate capacitor given the energy stored in the capacitor, the area of the plates, and the separation between the plates.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Separation between the plates, \( d = 2 \, \text{mm} = 0.002 \, \text{m} \)
- Area of the plates, \( A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \)
- Energy stored in the capacitor, \( U = 0.01 \, \text{J} \)

2. **Use the Formula for Electrostatic Force:**
The electrostatic force \( F \) between the plates of a capacitor can be expressed as:
\[ F = \frac{Q^2}{2 \epsilon_0 A} \]
where \( Q \) is the charge on the capacitor and \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

3. **Relate Energy to Charge and Capacitance:**
The energy stored in a capacitor is given by:
\[ U = \frac{Q^2}{2C} \]
where \( C \) is the capacitance. The capacitance for a parallel-plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]

4. **Substituting Capacitance into the Energy Equation:**
Rearranging the energy equation gives:
\[ Q^2 = 2U \cdot C \]
Substituting \( C \):
\[ Q^2 = 2U \cdot \frac{\epsilon_0 A}{d} \]

5. **Calculate \( Q^2 \):**
Plugging in the values:
\[ Q^2 = 2 \cdot 0.01 \cdot \frac{8.85 \times 10^{-12} \cdot 5 \times 10^{-4}}{0.002} \]
\[ Q^2 = 0.02 \cdot \frac{8.85 \times 10^{-12} \cdot 5 \times 10^{-4}}{0.002} \]
\[ Q^2 = 0.02 \cdot \frac{4.425 \times 10^{-15}}{0.002} \]
\[ Q^2 = 0.02 \cdot 2.2125 \times 10^{-12} = 4.425 \times 10^{-14} \, \text{C}^2 \]

6. **Substituting \( Q^2 \) into the Force Equation:**
Now substitute \( Q^2 \) back into the force equation:
\[ F = \frac{4.425 \times 10^{-14}}{2 \cdot 8.85 \times 10^{-12} \cdot 5 \times 10^{-4}} \]
\[ F = \frac{4.425 \times 10^{-14}}{8.85 \times 10^{-12} \cdot 10^{-3}} = \frac{4.425 \times 10^{-14}}{8.85 \times 10^{-15}} \approx 5 \, \text{N} \]

### Final Answer:
The electrostatic force of attraction between the plates is approximately **5 N**.