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Question

Three equal charges +q each, are placed at the three vertices of an equilateral triangle centred at the origin. They are held in equilibrium by a restoring force of magnitude F(r)=kr2 (where k is a constant, r is position of charge from origin) directed towards the origin. The distance of the any charge from the origin is (3q2pπϵ0k)14 where p=

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Solution



Here from coloumbs law, Fq=q24πϵ0a2
The net force acting on one of the charges is
Fnet=2Fqcos30=2q24πϵ0a232

position of charge from origin r=a3a=r3;
Hence Fnet=q24πϵ03r2.

For equilibrium F(r)=Fnetkr2=q24πϵ03r2r=(3q212πϵ0k)14

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