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JEE Main20231 Feb 2023Morning ShiftPhysicsElectrostaticsActual

Two equal positive point charges are separated by a distance 2 a . 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 ______.

Correct answer

0

Step-by-step solution

Let us assume at distance y , force is maximum. Now, force at a distance y is given by, F n e t = 2 F cos θ So, F n e t = 2 K q q 0 y y 2 + a 2 3 2 For F n e t to be maximum, d F n e t d y = 0 Or K q q 0 y 2 + a 2 3 2 - y 3 2 × 2 y y 2 + a 2 1 2 y 2 + a 2 3 = 0 Simplifying, we get y = a 2 . Hence, the value of x = 2 .

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