The potential at a point due to a charge of `4xx10^(-7)` C located 10 cm away is :

Options

A. `3.6 xx 10^5 V`
B. `3.6 xx 10^4 V`
C. `4.5 xx 10^4 V`
D. `4.5 xx 10^5 V`

Answer

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

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

where:
- \( V \) is the electric potential,
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (or \( \frac{1}{4 \pi \epsilon_0} \)),
- \( Q \) is the charge,
- \( r \) is the distance from the charge to the point where the potential is being calculated.

### Step-by-Step Solution:

1. **Identify the given values:**
- Charge \( Q = 4 \times 10^{-7} \, \text{C} \)
- Distance \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Convert the distance to meters:**
- Since \( r \) is given in centimeters, convert it to meters:
\[ r = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} = 0.1 \, \text{m} \]

3. **Substitute the values into the potential formula:**
- Using \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \):
\[ V = \frac{8.99 \times 10^9 \cdot (4 \times 10^{-7})}{0.1} \]

4. **Calculate the numerator:**
\[ 8.99 \times 10^9 \cdot 4 \times 10^{-7} = 35.96 \times 10^2 = 3.596 \times 10^3 \]

5. **Divide by the distance:**
\[ V = \frac{3.596 \times 10^3}{0.1} = 3.596 \times 10^4 \, \text{V} \]

6. **Final answer in standard form:**
\[ V = 3.6 \times 10^4 \, \text{V} \]

### Conclusion:
The potential at a point due to a charge of \( 4 \times 10^{-7} \, \text{C} \) located \( 10 \, \text{cm} \) away is \( 3.6 \times 10^4 \, \text{V} \).