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. \(8 \times 10^{-6}\,N\)
• B. \(8 \times 10^{-3}\,N\)
• C. \(8 \times 10^{-2}\,N\)
• D. \(8\,N\)

Hint:

To quickly solve electric field force problems: \[ F = qE \] Remember: \[ 1\,\mu C = 10^{-6} C \] Always convert microcoulombs to coulombs before calculation.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The goal is to calculate the electrostatic force acting on a point charge when it is subjected to a uniform electric field.
Step 2: Key Formula or Approach:
The force \(F\) on a charge \(q\) in an electric field \(E\) is defined by the formula:
\[ F = qE \]
Step 3: Detailed Explanation:
1. Given Data:
Charge, \(q = 2 \, \mu C = 2 \times 10^{-6} \, C\)
Electric Field, \(E = 4 \times 10^3 \, N/C\)
2. Calculation:
\[ F = (2 \times 10^{-6} \, C) \times (4 \times 10^3 \, N/C) \]
3. Simplification:
Multiply the coefficients: \(2 \times 4 = 8\).
Add the powers of 10: \((-6) + 3 = -3\).
\[ F = 8 \times 10^{-3} \, N \]
Step 4: Final Answer:
The force experienced by the charge is \(8 \times 10^{-3} \, N\).