Two electric charges of 9 `mu`C and -3 `mu`C are placed 0.16 m apart in air. There will be a point P at which electric potential is zero on the line joining two charges and in between them. The distance of P from 9 `mu`C is

Options

A. 0.14 m
B. 0.12 m
C. 0.08 m
D. 0.06 m

Answer

To solve the problem of finding the distance of point P from the charge of 9 µC where the electric potential is zero, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- Let charge \( Q_1 = 9 \, \mu C \) be at point A.
- Let charge \( Q_2 = -3 \, \mu C \) be at point B.
- The distance between the two charges is \( d = 0.16 \, m \).

2. **Define the Distance from Charge \( Q_1 \)**:
- Let the distance from charge \( Q_1 \) (9 µC) to point P be \( x \).
- Therefore, the distance from charge \( Q_2 \) (-3 µC) to point P will be \( 0.16 - x \).

3. **Write the Expression for Electric Potential**:
- The electric potential \( V \) at point P due to charge \( Q_1 \) is given by:
\[ V_1 = \frac{k \cdot Q_1}{x} = \frac{k \cdot 9 \times 10^{-6}}{x} \]
- The electric potential \( V \) at point P due to charge \( Q_2 \) is given by:
\[ V_2 = \frac{k \cdot Q_2}{0.16 - x} = \frac{k \cdot (-3 \times 10^{-6})}{0.16 - x} \]

4. **Set the Total Electric Potential to Zero**:
- For the total electric potential at point P to be zero:
\[ V_1 + V_2 = 0 \]
- This leads to:
\[ \frac{k \cdot 9 \times 10^{-6}}{x} + \frac{k \cdot (-3 \times 10^{-6})}{0.16 - x} = 0 \]

5. **Simplify the Equation**:
- We can cancel \( k \) (since it is a constant) from both sides:
\[ \frac{9 \times 10^{-6}}{x} = \frac{3 \times 10^{-6}}{0.16 - x} \]

6. **Cross-Multiply to Solve for x**:
- Cross-multiplying gives:
\[ 9(0.16 - x) = 3x \]
- Expanding this:
\[ 1.44 - 9x = 3x \]

7. **Combine Like Terms**:
- Bringing all terms involving \( x \) to one side:
\[ 1.44 = 12x \]

8. **Solve for x**:
- Dividing both sides by 12:
\[ x = \frac{1.44}{12} = 0.12 \, m \]

### Final Answer:
The distance of point P from the charge of 9 µC is **0.12 m**.