A positive test charge of 2.0×10^−4 C is placed in an electric field. The force on the test charge is 0.06 N. What is the electric field intensity at the location of the test charge?

Question

Group of answer choices

5.0 x 10^5 N/C

3.0 x 10^4 N/C

3.0 x 10^2 N/C

3.0×10^−5 N/C

Answer 1

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

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

where:

• \( E \) is the electric field intensity,
• \( F \) is the force on the charge,
• \( q \) is the charge.

Given:

• \( F = 0.06 , \text{N} \)
• \( q = 2.0 \times 10^{-4} , \text{C} \)

Now, we can plug the values into the formula:

\[ E = \frac{0.06 , \text{N}}{2.0 \times 10^{-4} , \text{C}} \]

Calculating this gives:

\[ E = \frac{0.06}{2.0 \times 10^{-4}} = 300 , \text{N/C} \]

or

\[ E = 3.0 \times 10^2 , \text{N/C} \]

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

3.0 x 10^2 N/C