The potential at a point due to charge of `5xx10^(-7)C` located 10 cm away is

Options

A. `3.5xx10^(5)V`
B. `3.5xx10^(4)V`
C. `4.5xx10^(4)V`
D. `4.5xx10^(5)V`

Explanation

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

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

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,
- \( R \) 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 \( Q = 5 \times 10^{-7} \, \text{C} \)
- Distance \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Convert the distance to meters:**
- Since \( 1 \, \text{cm} = 0.01 \, \text{m} \), we convert \( 10 \, \text{cm} \) to meters:
\[ R = 10 \, \text{cm} = 10 \times 0.01 = 0.1 \, \text{m} \]

3. **Substitute the values into the potential formula:**
\[ V = \frac{K \cdot Q}{R} \]
Plugging in the values:
\[ V = \frac{9 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 5 \times 10^{-7} \, \text{C}}{0.1 \, \text{m}} \]

4. **Calculate the numerator:**
\[ 9 \times 10^9 \cdot 5 \times 10^{-7} = 45 \times 10^2 = 4.5 \times 10^3 \]

5. **Now divide by the distance:**
\[ V = \frac{4.5 \times 10^3}{0.1} = 4.5 \times 10^4 \, \text{V} \]

6. **Final Result:**
The potential at the point due to the charge is:
\[ V = 4.5 \times 10^4 \, \text{V} \]

### Answer:
The potential at a point due to a charge of \( 5 \times 10^{-7} \, \text{C} \) located \( 10 \, \text{cm} \) away is \( 4.5 \times 10^4 \, \text{V} \).

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