Electrostatics

JEE Main level, single correct, medium

165 of 311

Question 165 of 311

Consider a circular loop that is uniformly charged and has a radius a2\mathrm{a} \sqrt{2}. Find the position along the positive z -axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in xy-plane at the origin :

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Answer: (C) a

E=KQr(x2+R2)3/2dEdx=0∴x=R2=2a2=a\begin{aligned} & \mathrm{E}=\frac{\mathrm{KQr}}{\left(\mathrm{x}^2+\mathrm{R}^2\right)^{3 / 2}} \\ & \frac{\mathrm{dE}}{\mathrm{dx}}=0 \\ & \therefore \mathrm{x}=\frac{\mathrm{R}}{\sqrt{2}}=\frac{\sqrt{2} \mathrm{a}}{\sqrt{2}}=\mathrm{a}\end{aligned}

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