Question

What is the electric potential at a distance of 1.2 m from a 7.5 µC point charge?

5.6 × 104 V
8.1 × 104 V
5.6 × 1010 V
8.1 × 1010 V

Answer 1

The electric potential \( V \) due to a point charge \( Q \) at a distance \( r \) is given by the formula:

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

where:
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( Q \) is the charge in coulombs,
- \( r \) is the distance in meters.

Given:
- \( Q = 7.5 \, \mu\text{C} = 7.5 \times 10^{-6} \, \text{C} \)
- \( r = 1.2 \, \text{m} \)

Now, plug in the values into the formula:

\[
V = \frac{(8.99 \times 10^9) \cdot (7.5 \times 10^{-6})}{1.2}
\]

Calculating the numerator:

\[
8.99 \times 10^9 \cdot 7.5 \times 10^{-6} = 67.425 \times 10^3 = 67425 \, \text{V m}
\]

Now divide by \( r \):

\[
V = \frac{67425 \, \text{V m}}{1.2} \approx 56187.5 \, \text{V}
\]

Thus,

\[
V \approx 5.619 \times 10^4 \, \text{V}
\]

This is approximately equal to \( 5.6 \times 10^4 \, \text{V} \).

The correct answer is \( \boxed{5.6 \times 10^4 \, \text{V}} \).