Question

A point charge of 8.0 C is placed 0.050 m from a rod that has an electric field of 1.5 × 103 N/C.

What is the electric potential energy of the point charge?

6.0 × 10-4 J
2.4 × 10-1 J
6.0 × 102 J
2.4 × 105 J

Answer 1

To find the electric potential energy \( U \) of a point charge in an electric field, you can use the formula:

\[
U = q \cdot E \cdot d
\]

where:

- \( U \) is the electric potential energy,
- \( q \) is the charge (8.0 C in this case),
- \( E \) is the electric field (1.5 × 10³ N/C), and
- \( d \) is the distance from the charge to the source of the electric field (0.050 m).

Now, substituting the values into the formula:

\[
U = (8.0 \, \text{C}) \cdot (1.5 \times 10^3 \, \text{N/C}) \cdot (0.050 \, \text{m})
\]

Now, calculate it step by step:

1. Calculate \( E \cdot d \):

\[
E \cdot d = (1.5 \times 10^3 \, \text{N/C}) \cdot (0.050 \, \text{m}) = 75 \, \text{V}
\]

2. Now, multiply by the charge \( q \):

\[
U = 8.0 \, \text{C} \cdot 75 \, \text{V} = 600 \, \text{J}
\]

Thus, the electric potential energy of the point charge is:

\[
U = 600 \, \text{J} = 6.0 \times 10^2 \, \text{J}
\]

So, the correct answer is:

**6.0 × 10² J**