What is the electric potential at a distance of 9 cm from `3 mu C`?

Options

A. 3 V
B. 300 V
C. 270 V
D. 3x`10^5` V

Answer

To find the electric potential at a distance of 9 cm from a point charge of 3 µC, 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 (\( k \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( Q \) is the charge (in coulombs),
- \( r \) is the distance from the charge (in meters).

### Step-by-step Solution:

1. **Convert the charge from microcoulombs to coulombs:**
\[ Q = 3 \, \mu C = 3 \times 10^{-6} \, C \]

2. **Convert the distance from centimeters to meters:**
\[ r = 9 \, cm = 9 \times 10^{-2} \, m \]

3. **Substitute the values into the formula:**
\[ V = \frac{(9 \times 10^9) \cdot (3 \times 10^{-6})}{9 \times 10^{-2}} \]

4. **Calculate the numerator:**
\[ 9 \times 10^9 \cdot 3 \times 10^{-6} = 27 \times 10^3 = 2.7 \times 10^4 \, \text{(after adjusting the powers of ten)} \]

5. **Now divide by the distance:**
\[ V = \frac{2.7 \times 10^4}{9 \times 10^{-2}} = \frac{2.7 \times 10^4}{0.09} \]

6. **Perform the division:**
\[ V = 3 \times 10^5 \, \text{volts} \]

### Final Answer:
The electric potential at a distance of 9 cm from a charge of 3 µC is:
\[ V = 3 \times 10^5 \, \text{volts} \]