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

Concept: The force experienced by a charge placed in an electric field is given by: \[ F = qE \] where
• \(F\) = Electric force
• \(q\) = Charge
• \(E\) = Electric field intensity

Step 1: Write the given quantities. \[ q = 2\,\mu C = 2 \times 10^{-6} C \] \[ E = 4 \times 10^{3} \, N/C \]

Step 2: Apply the formula \(F = qE\). \[ F = (2 \times 10^{-6})(4 \times 10^{3}) \]

Step 3: Calculate the force. \[ F = 8 \times 10^{-3} \, N \] \[ \boxed{F = 8 \times 10^{-3} \, N} \]