The 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 of finding the electrostatic force of attraction between the plates of a parallel-plate capacitor, we can follow these steps:

### Step 1: Convert the given values to SI units
- The separation between the plates \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- The area of the plates \( A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \)
- The energy stored in the capacitor \( U = 0.01 \, \text{J} = 10^{-2} \, \text{J} \)

### Step 2: Use the formula for electrostatic force
The electrostatic force \( F \) between the plates of the capacitor can be expressed in terms of the energy stored and the separation distance:
\[ F = \frac{U}{d} \]

### Step 3: Substitute the known values into the formula
Now, substituting the values of \( U \) and \( d \):
\[ F = \frac{10^{-2} \, \text{J}}{2 \times 10^{-3} \, \text{m}} = \frac{10^{-2}}{2 \times 10^{-3}} = \frac{10^{-2}}{2 \times 10^{-3}} = 5 \, \text{N} \]

### Step 4: Conclusion
Thus, the electrostatic force of attraction between the plates is:
\[ \boxed{5 \, \text{N}} \]