Electrostatics

JEE Main level, single correct, medium

179 of 311

Question 179 of 311

Electric potential at a point PP due to a point charge of 5×10−9C5\times {10}^{-9}C is 50V50V. The distance of PP from the point charge is:
(Assume, 14πϵ0=9×109Nm2C−2\dfrac{1}{4\pi {\epsilon }_{0}}=9\times {10}^{9}N{m}^{2}{C}^{-2})

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Answer: (D) 90cm90cm

The electric potential for a point charge is given by
V=Q4πϵ0rV=\dfrac{Q}{4\pi {\epsilon }_{0}r}
The data given is
V=50VQ=5×10−9CV=50V Q=5\times {10}^{-9}C
So, the distance of the point is
r=14πϵ0QVr=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{Q}{V}
⇒r=9×109×5×10−950\Rightarrow r=9\times {10}^{9}\times \dfrac{5\times {10}^{-9}}{50}
⇒r=0.9m=90cm\Rightarrow r=0.9m=90cm

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