Question

A positive test charge of 1.0×10−6 C is placed in an electric field. The force on the test charge is 0.04 N. What is the electric field intensity at the location of the test charge?

Group of answer choices

4.0 x 105 N/C

4.0 x 104 N/C

3.0×10−5 N/C

6.0 x 104 N/C

Answer 1

To find the electric field intensity (E) at the location of the test charge, you can use the formula:

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

where:
- \( E \) is the electric field intensity,
- \( F \) is the force on the test charge,
- \( q \) is the charge.

Given:
- \( F = 0.04 \, \text{N} \)
- \( q = 1.0 \times 10^{-6} \, \text{C} \)

Now, plug in the values:

\[
E = \frac{0.04 \, \text{N}}{1.0 \times 10^{-6} \, \text{C}} = \frac{0.04}{1.0 \times 10^{-6}} = 4.0 \times 10^4 \, \text{N/C}
\]

Therefore, the electric field intensity at the location of the test charge is:

\[
\boxed{4.0 \times 10^4 \, \text{N/C}}
\]