Question

A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of \(6 \times 10^4\) N/C. Then the magnitude of the charge is

• A. \(5 \times 10^{-4}\) C
• B. \(5 \times 10^{-10}\) C
• C. \(-18 \times 10^{-6}\) C
• D. \(-5 \times 10^{-9}\) C

Hint:

For stationary particle, \(qE = mg\), direction of E determines sign.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Electric force balances weight: \(qE = mg\).
Step 2: Detailed Explanation:
\(m = 0.003\ \mathrm{g} = 3 \times 10^{-6}\ \mathrm{kg}\)
\(q = mg/E = (3 \times 10^{-6} \times 10)/(6 \times 10^4) = 5 \times 10^{-10}\ \mathrm{C}\)
Step 3: Final Answer:
Charge magnitude is \(5 \times 10^{-10}\) C.