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Discussion Response Post to: H.C.verma question

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22 May 2008 20:49:45 IST

Subject: Re:H.C.verma question

sriram.a (222)

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31)yaar simple take a charge at perpendicular bisector at a distance x.

now resolve the components.The forces perpendicular to bisector cancels.

$F=\frac{kQqx}{\sqrt{(d^2/4+x^2})^3}$ .

now for force to be maximum $\frac{dF}{dx}=0$ .

solve it x= $\frac{d}{2\sqrt{2}}$ .now substitute in F to get maximum force maximum.

<SRIRAM.A> on high way of IIT

this reply: 2 points (with 0 Olaaa!! Perrrfect answer.

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