Electrostatics

JEE Main level, single correct, medium

200 of 311

Question 200 of 311

A system of three charges are placed as shown in the figure:

If D>>d,D\gt \gt d, the potential energy of the system is best given by:

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Answer: (A) 14πϵ0[−q2d−qQdD2]\dfrac{1}{4\pi {ϵ}_{0}}\left[-\dfrac{{q}^{2}}{d}-\dfrac{qQd}{{D}^{2}}\right]

U=k(−q)(q)d+k(−q)(Q)(d2−D)+k(q)(Q)(D+d2)U=\dfrac{k\left(-q\right)\left(q\right)}{d}+\dfrac{k\left(-q\right)\left(Q\right)}{\left(\dfrac{d}{2}-D\right)}+\dfrac{k\left(q\right)\left(Q\right)}{\left(D+\dfrac{d}{2}\right)}
=−kq2d−kQq(D−d2)+KQq(D+d2)=\dfrac{-k{q}^{2}}{d}-\dfrac{kQq}{\left(D-\dfrac{d}{2}\right)}+\dfrac{KQq}{\left(D+\dfrac{d}{2}\right)}
=−kq2d+kQq(−D−d2+D−d2D2−d24)=\dfrac{-k{q}^{2}}{d}+kQq\left(\dfrac{-D-\dfrac{d}{2}+D-\dfrac{d}{2}}{{D}^{2}-\dfrac{{d}^{2}}{4}}\right)
=−kq2d−kQqdD2−d24=\dfrac{-k{q}^{2}}{d}-\dfrac{kQqd}{{D}^{2}-\dfrac{{d}^{2}}{4}}
Since, D>>d
U=−kq2d−kQqdD2=14πϵ0[−q2d−qQdD2]U=\dfrac{-k{q}^{2}}{d}-\dfrac{kQqd}{{D}^{2}}=\dfrac{1}{4\pi {ϵ}_{0}}\left[-\dfrac{{q}^{2}}{d}-\dfrac{qQd}{{D}^{2}}\right]

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