Question:

A conducting sphere of radius 5 cm has an unknown charge. The electric field at 10 cm from the center of the sphere is $1.8 \times 10^3$ N/C. The net charge on the sphere is

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Use $E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q}{r^2}$ for points outside a uniformly charged conducting sphere.
Updated On: May 12, 2025
  • 1.8 nC
  • 2 nC
  • 1.5 nC
  • 1.2 nC
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The Correct Option is B

Solution and Explanation

Electric field outside a conducting sphere: $E = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q}{r^2}$
Rewriting: $q = E \cdot r^2 \cdot 4\pi\varepsilon_0$
$E = 1.8 \times 10^3$ N/C, $r = 10$ cm = 0.1 m
$\varepsilon_0 = 8.85 \times 10^{-12}$ C$^2$/Nm$^2$
$q = 1.8 \times 10^3 \cdot (0.1)^2 \cdot 4\pi \cdot 8.85 \times 10^{-12}$
$q = 1.8 \cdot 0.01 \cdot 4\pi \cdot 8.85 \times 10^{-12} \approx 2 \times 10^{-9}$ C = 2 nC
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