Question

A negative charge of 8.0×10−7 C experiences a force of 0.03 N to the left in an electric field. What is the magnitude of the field?
Group of answer choices

2.5 x 10-2 N/C

3.5 x 105 N/C

2.5 x 102 N/C

3.75 x 104 N/C

Answer 1

To find the magnitude of the electric field (E), we can use the formula:

\[
F = qE
\]

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

Rearranging this formula to solve for the electric field \( E \):

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

Given:
- \( F = 0.03 \, \text{N} \)
- \( q = -8.0 \times 10^{-7} \, \text{C} \)

Now, substituting the values into the formula:

\[
E = \frac{0.03 \, \text{N}}{|-8.0 \times 10^{-7} \, \text{C}|} = \frac{0.03 \, \text{N}}{8.0 \times 10^{-7} \, \text{C}}
\]

Calculating this gives:

\[
E = \frac{0.03}{8.0 \times 10^{-7}} = 3.75 \times 10^4 \, \text{N/C}
\]

So, the magnitude of the electric field is:

\[
E = 3.75 \times 10^4 \, \text{N/C}
\]

Thus, the correct choice from the options provided is:

**3.75 x 10^4 N/C**.