Published by:
CGP EDU Academic Team
Published on: September 12, 2026
What is the magnitude of a point charge which produces an electric field of $2\mathrm{N}/\mathrm{coulomb}$ at a distance of $60 \text{ cm} \left( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ N} - \text{m}^2 / \text{C}^2 \right)$
Text Solution
Verified by ExpertsThe correct answer is:
A
$E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$
⇒ ⇒ $2 = 9 \times 10^{9} \times \frac{Q}{(0.6)^{2}}$ ⇒ ⇒ Q = 8 × × $10^{-11} C$
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