Two electric charges of `9mu `C and `- 3mu c` are placed 0.16 m apart in air. There are two points A and B on the line joining the two charges at distances of (i) 0.04 m from-3`mu`C and in between the charges and (ii) 0.08m from - 3`mu `C and outside the two charges. The potentials at A and B

Options

A. `0V,0V`
B. `IV ,2V`
C. `3 V ,4V`
D. `IV ,7V`

Answer

To solve the problem, we need to calculate the electric potential at two points A and B due to the two charges: \( q_1 = 9 \, \mu C \) and \( q_2 = -3 \, \mu C \), which are separated by a distance of \( 0.16 \, m \).

### Step 1: Understand the Position of Charges and Points
- Let \( q_1 = 9 \, \mu C \) be located at point \( O \) (0 m).
- Let \( q_2 = -3 \, \mu C \) be located at point \( P \) (0.16 m).
- Point A is \( 0.04 \, m \) from \( q_2 \) and is located between the two charges, which means it is \( 0.12 \, m \) from \( q_1 \).
- Point B is \( 0.08 \, m \) from \( q_2 \) and is located outside the two charges.

### Step 2: Calculate the Potential at Point A
The electric potential \( V \) at a point due to a charge \( q \) is given by the formula:
\[ V = k \frac{q}{r} \]
where \( k = 9 \times 10^9 \, Nm^2/C^2 \) (Coulomb's constant), \( q \) is the charge, and \( r \) is the distance from the charge to the point.

1. **Calculate the potential at A due to \( q_1 \)**:
- Distance from \( q_1 \) to A: \( r_1 = 0.12 \, m \)
- Potential due to \( q_1 \):
\[ V_{A1} = k \frac{q_1}{r_1} = 9 \times 10^9 \frac{9 \times 10^{-6}}{0.12} \]

2. **Calculate the potential at A due to \( q_2 \)**:
- Distance from \( q_2 \) to A: \( r_2 = 0.04 \, m \)
- Potential due to \( q_2 \):
\[ V_{A2} = k \frac{q_2}{r_2} = 9 \times 10^9 \frac{-3 \times 10^{-6}}{0.04} \]

3. **Total potential at A**:
\[ V_A = V_{A1} + V_{A2} \]
Substitute the values:
\[ V_A = 9 \times 10^9 \left( \frac{9 \times 10^{-6}}{0.12} - \frac{3 \times 10^{-6}}{0.04} \right) \]

### Step 3: Calculate the Potential at Point B
1. **Calculate the potential at B due to \( q_1 \)**:
- Distance from \( q_1 \) to B: \( r_3 = 0.16 + 0.08 = 0.24 \, m \)
- Potential due to \( q_1 \):
\[ V_{B1} = k \frac{q_1}{r_3} = 9 \times 10^9 \frac{9 \times 10^{-6}}{0.24} \]

2. **Calculate the potential at B due to \( q_2 \)**:
- Distance from \( q_2 \) to B: \( r_4 = 0.08 \, m \)
- Potential due to \( q_2 \):
\[ V_{B2} = k \frac{q_2}{r_4} = 9 \times 10^9 \frac{-3 \times 10^{-6}}{0.08} \]

3. **Total potential at B**:
\[ V_B = V_{B1} + V_{B2} \]
Substitute the values:
\[ V_B = 9 \times 10^9 \left( \frac{9 \times 10^{-6}}{0.24} - \frac{3 \times 10^{-6}}{0.08} \right) \]

### Step 4: Simplify and Calculate
1. **For Point A**:
\[ V_A = 9 \times 10^9 \left( \frac{9 \times 10^{-6}}{0.12} - \frac{3 \times 10^{-6}}{0.04} \right) = 0 \, V \]

2. **For Point B**:
\[ V_B = 9 \times 10^9 \left( \frac{9 \times 10^{-6}}{0.24} - \frac{3 \times 10^{-6}}{0.08} \right) = 0 \, V \]

### Final Answer
- The potential at point A is \( 0 \, V \).
- The potential at point B is \( 0 \, V \).