Question

A negative charge of 4.0×10−4 C experiences a force of 0.2 N to the right in an electric field. What is the magnitude of the field?

Group of answer choices

2.5 x 102 N/C

2.5 x 10-2 N/C

5 x 102 N/C

5 x 104 N/C

Answer 1

To find the magnitude of the electric field (E) experienced by a charge (q), we can use the formula:

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

where:
- \( E \) is the electric field,
- \( F \) is the force experienced by the charge,
- \( q \) is the magnitude of the charge.

Given:
- \( F = 0.2 \, \text{N} \)
- \( q = 4.0 \times 10^{-4} \, \text{C} \)

Substituting the values into the formula:

\[
E = \frac{0.2 \, \text{N}}{4.0 \times 10^{-4} \, \text{C}}
\]

Calculating:

\[
E = \frac{0.2}{4.0 \times 10^{-4}} = 0.2 \div 0.0004 = 500 \, \text{N/C}
\]

In scientific notation, this is:

\[
E = 5.0 \times 10^{2} \, \text{N/C}
\]

Thus, the magnitude of the electric field is \( 5 \times 10^{2} \, \text{N/C} \).

The correct answer from the choices provided is:

**5 x 10² N/C**