Two equal positive point charges are separated by a distance 2a. The distance of a point from the centre of the line joining two charges on the…

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Two equal positive point charges are separated by a distance 2a. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge q_0 becomes maximum is a√x. The value of x is ______.

MARKS Content Team · Accepted Answer

Correct answer: 2. Let us assume at distance y, force is maximum. Now, force at a distance y is given by, F_net=2Fcosθ So, F_net=(2Kqq_0yy2+a2^32For F_net to be maximum, dF_net)/dy=0 Or Kqq_0y2+a2^32-y3/2×2yy2+a2^12y2+a2^3=0 Simplifying, we get y=a√2. Hence, the value of x=2.

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Two equal positive point charges are separated by a distance 2a. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge q0 becomes maximum is ax. The value of x is ______.

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