The electrostatic force on a small sphere of charge `0.4 muC` due to another small sphere of charge `-0.8 mu C` in air 0.2 N (i) What is the distance between the two spheres? (ii) What is the force on the second sphere due to the first ?

Options

A. The distance between the two spheres is 0.12 m. `0.2 N`
B. The distance between the two spheres is 0.12 m. `1.2 N`
C. The distance between the two spheres is 0.12 m. `2.2 N`
D. The distance between the two spheres is 0.22 m. `22.2 N`

Explanation

To solve the problem step by step, we will use Coulomb's law, which states that the electrostatic force \( F \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( R \) is given by:

\[ F = \frac{1}{4\pi\epsilon_0} \cdot \frac{|Q_1 Q_2|}{R^2} \]

Where:
- \( F \) is the magnitude of the force between the charges,
- \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges,
- \( R \) is the distance between the charges,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Part (i): Finding the distance between the two spheres

1. **Identify the given values**:
- \( Q_1 = 0.4 \, \mu C = 0.4 \times 10^{-6} \, C \)
- \( Q_2 = -0.8 \, \mu C = -0.8 \times 10^{-6} \, C \)
- \( F = 0.2 \, N \)
- The constant \( \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

2. **Set up the equation using Coulomb's law**:
\[ F = \frac{1}{4\pi\epsilon_0} \cdot \frac{|Q_1 Q_2|}{R^2} \]
Substituting the known values:
\[ 0.2 = 9 \times 10^9 \cdot \frac{|0.4 \times 10^{-6} \cdot (-0.8 \times 10^{-6})|}{R^2} \]

3. **Calculate the product of the charges**:
\[ |Q_1 Q_2| = |0.4 \times 10^{-6} \cdot -0.8 \times 10^{-6}| = 0.32 \times 10^{-12} \, C^2 \]

4. **Rearranging the equation to solve for \( R^2 \)**:
\[ R^2 = \frac{9 \times 10^9 \cdot 0.32 \times 10^{-12}}{0.2} \]

5. **Calculate \( R^2 \)**:
\[ R^2 = \frac{2.88 \times 10^{-3}}{0.2} = 1.44 \times 10^{-2} \]

6. **Taking the square root to find \( R \)**:
\[ R = \sqrt{1.44 \times 10^{-2}} = 0.12 \, m = 12 \, cm \]

### Part (ii): Finding the force on the second sphere due to the first sphere

1. **Use Newton's Third Law**:
According to Newton's Third Law, the force on the second sphere due to the first sphere \( F_{21} \) is equal in magnitude and opposite in direction to the force on the first sphere due to the second sphere \( F_{12} \).

2. **Thus, we have**:
\[ F_{21} = -F_{12} \]
Since \( F_{12} = 0.2 \, N \), we have:
\[ F_{21} = 0.2 \, N \]

### Final Answers:
- (i) The distance between the two spheres is \( 12 \, cm \).
- (ii) The force on the second sphere due to the first sphere is \( 0.2 \, N \).