Published by:
CGP EDU Academic Team
Published on: September 12, 2026
What is the magnitude of a point charge due to which the electric field $30 \mathrm{cm}$ away has the magnitude 2 newton / coulomb $\left[ \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^{9} \, \mathrm{Nm^{2} / C^{2}} \right]$
Text Solution
Verified by ExpertsThe correct answer is:
A
Electric field due to a point charge $E = \frac{q}{4 \pi \varepsilon_0 r^2}$
∴ ∴ $q = E \times 4 \pi \varepsilon_0 r^2 = 2 \times \frac{1}{9 \times 10^9} \times \left(\frac{30}{100}\right)^2$ = 2 × × 10 –11 C
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