Question

A charge of +0.08 C moves to the right due to a 4 N force exerted by an electric field. What is the magnitude and direction of the electric field?

0.32 N/C left
0.32 N/C right
50 N/C left
50 N/C right

Answer 1

To find the magnitude and direction of the electric field, we can use the formula that relates force, charge, and electric field:

\[
F = qE
\]

Where:
- \( F \) is the force (in Newtons),
- \( q \) is the charge (in Coulombs),
- \( E \) is the electric field (in N/C).

Given:
- \( F = 4 \, \text{N} \) (to the right),
- \( q = +0.08 \, \text{C} \).

We can rearrange the formula to solve for the electric field \( E \):

\[
E = \frac{F}{q}
\]

Substituting the values into the equation:

\[
E = \frac{4 \, \text{N}}{0.08 \, \text{C}} = 50 \, \text{N/C}
\]

Since the force is directed to the right and the charge is positive, the electric field must also be directed to the right.

Thus, the magnitude and direction of the electric field is:

\[
\text{50 N/C right}
\]

So the correct answer is **50 N/C right**.