Electrostatics

JEE Main level, single correct, medium

207 of 311

Question 207 of 311

Charges −q-q and +q+q, located at AA and BB, respectively, constitute an electric dipole. Distance AB=2aAB=2a, OO is the mid point of the dipole and OPOP is perpendicular to ABAB. A charge QQ is placed at PP where OP=yOP=y and y≫2ay≫2a. The charge QQ experiences an electrostatic force FF. If QQ is now moved along the equatorial line to P′P' such that OP′=(y3)OP'=\left(\dfrac{y}{3}\right) the force on QQ will be close to(y3≪2a)\left(\dfrac{y}{3}≪2a\right)

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Answer: (A) 27F27F.

Electric field intensity at the distance yy along the perpendicular bisector is

∣E1∣=KPy3\left|{E}_{1}\right|=\dfrac{KP}{{y}^{3}}

So, force at distance yy on QQ,

F=Q(E1)=Q(KPy3)F=Q({E}_{1})=Q\left(\dfrac{KP}{{y}^{3}}\right) ...(1)(1)

F′=QKP(y3)3F'=Q\dfrac{KP}{{\left(\dfrac{y}{3}\right)}^{3}} ...(2)(2)
Dividing (1)\left(1\right) by (2)\left(2\right), we get,
F′F=(3)3\dfrac{F'}{F}={\left(3\right)}^{3}

∴F′=27F∴F'=27F

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