Two charges -4 `mu`C and 8 `mu`C are 27 cm apart. The distance from the first charge on line of joining between two charges where electric potential would be zero is

Options

A. 10 cm
B. 9 cm
C. 20 cm
D. 21 cm

Answer

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

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
- Let \( Q_1 = -4 \, \mu C = -4 \times 10^{-6} \, C \) (first charge)
- Let \( Q_2 = 8 \, \mu C = 8 \times 10^{-6} \, C \) (second charge)
- The distance between the two charges is \( R = 27 \, cm = 0.27 \, m \).

2. **Set Up the Problem**:
- We need to find a point \( P \) on the line joining the two charges where the electric potential \( V \) is zero.
- Let the distance from the first charge \( Q_1 \) to point \( P \) be \( x \). Therefore, the distance from the second charge \( Q_2 \) to point \( P \) will be \( R - x \).

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

4. **Set the Total Potential to Zero**:
- For the total potential at point \( P \) to be zero, we have:
\[ V_1 + V_2 = 0 \]
- Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ \frac{k Q_1}{x} + \frac{k Q_2}{R - x} = 0 \]

5. **Simplify the Equation**:
- We can cancel \( k \) from both sides:
\[ \frac{Q_1}{x} + \frac{Q_2}{R - x} = 0 \]
- Rearranging gives:
\[ \frac{Q_1}{x} = -\frac{Q_2}{R - x} \]

6. **Substitute the Values of Charges**:
- Substitute \( Q_1 = -4 \times 10^{-6} \, C \) and \( Q_2 = 8 \times 10^{-6} \, C \):
\[ \frac{-4 \times 10^{-6}}{x} = -\frac{8 \times 10^{-6}}{27 - x} \]

7. **Cross-Multiply and Solve for \( x \)**:
- Cross-multiplying gives:
\[ -4 \times 10^{-6} (27 - x) = -8 \times 10^{-6} x \]
- Simplifying:
\[ 4(27 - x) = 8x \]
\[ 108 - 4x = 8x \]
\[ 108 = 12x \]
\[ x = \frac{108}{12} = 9 \, cm \]

8. **Conclusion**:
- The distance from the first charge where the electric potential is zero is \( x = 9 \, cm \).