A negative charge of 2.0×10−3 C experiences a force of 0.03 N to the right in an electric field. What is the magnitude of the field?

Question

Group of answer choices

1.5 x 101 N/C

2.5 x 102 N/C

2.5 x 10-2 N/C

3.5 x 104 N/C

Answer 1

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

\[ F = qE \]

Where:

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

Given:

• \(F = 0.03 , \text{N}\)
• \(q = -2.0 \times 10^{-3} , \text{C}\) (the negative sign indicates the charge is negative, but it doesn't affect the magnitude of the electric field)

We need to solve for \(E\):

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

Substituting in the given values:

\[ E = \frac{0.03 , \text{N}}{2.0 \times 10^{-3} , \text{C}} = \frac{0.03}{0.002} = 15 , \text{N/C} \]

Now we look at the options provided:

1. \(1.5 \times 10^1 , \text{N/C}\) (which is equal to 15 N/C)
2. \(2.5 \times 10^2 , \text{N/C}\) (which is 250 N/C)
3. \(2.5 \times 10^{-2} , \text{N/C}\) (which is 0.025 N/C)
4. \(3.5 \times 10^4 , \text{N/C}\) (which is 35000 N/C)

The correct answer is:

1.5 x 10^1 N/C (15 N/C)