The potential at a point, due to a positive charge of `100 mu C` at a distance of `9 m`, is

Options

A. `10^(4) V`
B. `10^(5) V`
C. `10^(6) V`
D. `10^(7) V`

Explanation

To find the electric potential at a point due to a positive charge, we can use the formula for electric potential \( V \) due to a point charge:

\[ V = \frac{k \cdot Q}{R} \]

where:
- \( V \) is the electric potential,
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( Q \) is the charge (in coulombs),
- \( R \) is the distance from the charge to the point where the potential is being measured (in meters).

### Step-by-Step Solution:

1. **Identify the given values**:
- Charge \( Q = 100 \, \mu C = 100 \times 10^{-6} \, C \)
- Distance \( R = 9 \, m \)

2. **Substitute the values into the formula**:
\[ V = \frac{(9 \times 10^9) \cdot (100 \times 10^{-6})}{9} \]

3. **Simplify the equation**:
- First, calculate the numerator:
\[ 9 \times 10^9 \cdot 100 \times 10^{-6} = 9 \times 100 \times 10^3 = 900 \times 10^3 \]
- Now, substitute this back into the equation:
\[ V = \frac{900 \times 10^3}{9} \]

4. **Perform the division**:
\[ V = 100 \times 10^3 = 10^5 \, V \]

5. **Final result**:
\[ V = 10^5 \, V = 100000 \, V \]

### Conclusion:
The potential at a point due to a positive charge of \( 100 \, \mu C \) at a distance of \( 9 \, m \) is \( 100000 \, V \) or \( 10^5 \, V \).