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Question

Three equal charges each +q are placed (fixed) on the corners of an equilateral triangle of side a. Then, the Coulomb force experienced by one charge due to the rest of the two is :

A
14πε0q2a2
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B
14πε02q2a2
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C
14πε03q2a2
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D
zero
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Solution

The correct option is C 14πε03q2a2

Generally coulomb force between two charges q1 and q2 is F=q1q24πϵ0r3r where r is the distance between the charges.

From figure, BC=a2a24=32a

Let FO be the coulomb force on charge at O due to rest charges at A(a,0) and B(a2,32a)

thus, FO=q24πϵ0a3(a^i)+q24πϵ0a3(a2^i+3a2^j)=q24πϵ0a2(32^i+32^j)

|FO|=(q24πϵ0a2)2(94+34)=14πϵ0.3q2a2


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