Two positive charges of magnitude `4 C` and `6 C` are placed `10 cm` apart . The electric potential at a distance of `10 cm` from the middle point on the right bisector of the line,joining the two charges is

Options

A. `5 xx 10^(11) V`
B. `4 xx 10^(9) V`
C. `8 xx 10^(11) V`
D. `5 xx 10^(9) V`

Explanation

To solve the problem of finding the electric potential at a point located 10 cm from the midpoint on the right bisector of the line joining two positive charges of magnitudes 4 C and 6 C, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have two charges: \( Q_1 = 4 \, C \) and \( Q_2 = 6 \, C \).
- The distance between the two charges is \( d = 10 \, cm = 0.1 \, m \).
- The midpoint between the charges is at a distance of \( 5 \, cm = 0.05 \, m \) from each charge.
- We need to find the electric potential at a point \( P \) that is \( 10 \, cm \) away from the midpoint on the right bisector.

2. **Calculating the Distance from Each Charge to Point P**:
- The distance from the midpoint to point \( P \) is \( 10 \, cm = 0.1 \, m \).
- The distance from \( Q_1 \) (4 C) to point \( P \) is:
\[ R_1 = \sqrt{(0.05)^2 + (0.1)^2} = \sqrt{0.0025 + 0.01} = \sqrt{0.0125} = 0.1118 \, m \]
- The distance from \( Q_2 \) (6 C) to point \( P \) is:
\[ R_2 = \sqrt{(0.05)^2 + (0.1)^2} = \sqrt{0.0125} = 0.1118 \, m \]

3. **Using the Formula for Electric Potential**:
- The electric potential \( V \) at a point due to a point charge is given by:
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \]
- The total electric potential at point \( P \) due to both charges is:
\[ V_P = V_1 + V_2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1}{R_1} + \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_2}{R_2} \]
- Substituting the values:
\[ V_P = \frac{1}{4\pi \epsilon_0} \left( \frac{4}{0.1118} + \frac{6}{0.1118} \right) \]
- Simplifying:
\[ V_P = \frac{1}{4\pi \epsilon_0} \cdot \frac{10}{0.1118} \]

4. **Calculating the Numerical Value**:
- Using \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \):
\[ V_P = \frac{9 \times 10^9}{0.1118} \cdot 10 \]
- Calculating:
\[ V_P \approx 9 \times 10^9 \cdot 89.4 \approx 8.05 \times 10^{11} \, V \]

5. **Final Result**:
- The electric potential at point \( P \) is approximately \( 8.05 \times 10^{11} \, V \).