Question

A charge of 2 $\mu C$ is placed in an electric field of intensity \( 4 \times 10^3 \, \text{N/C} \). What is the force experienced by the charge?

• A. \( 8 \times 10^{-3} \, \text{N} \)
• B. \( 8 \times 10^{-6} \, \text{N} \)
• C. \( 4 \times 10^{-3} \, \text{N} \)
• D. \( 4 \times 10^{-6} \, \text{N} \)

Hint:

To calculate the force in an electric field, use the formula \( F = qE \), where \( q \) is the charge and \( E \) is the electric field intensity.

Answer 1

The Correct Option is A

The formula for the force \( F \) experienced by a charge \( q \) in an electric field \( E \) is \( F = qE \). Given: - \( q = 2 \, \mu C = 2 \times 10^{-6} \, C \) - \( E = 4 \times 10^3 \, \text{N/C} \) Substituting these values into the formula: \[ F = (2 \times 10^{-6}) \times (4 \times 10^3) = 8 \times 10^{-3} \, \text{N} \] The resulting force on the charge is \( 8 \times 10^{-3} \, \text{N} \).