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Question

# Four charges, each equal to -Q, are placed at the corners of a square and a charge +q placed at its centre. If the system is in equilibrium, the value of q is

A

Q4(1+22)

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B

Q4(1+22)

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C

Q2(1+22)

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D

Q2(1+22)

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Solution

## The correct option is A Q4(1+2√2) Let the side of the square be a OA=OC =r =a√2 1 Stability of charge +q at the centre Charges - Q at corners A and C will attract charge + q with equal and opposite forces. Similarly charges -Q at corners B and D will attract charge + q with equal and opposite force. Hence no net force acts on charge -q. 2. Stability of charge -Q at any corner Let us find the forces on charge - Q at corner A. This charge will experience four forces: (i) Force of repulsion Fi due to charge - Q at B (ii) Force of repulsion F2 due to charge - Q at D (iii) Force of repulsion F3 due to charge - Q at C (iv) Force of attraction F' due to charge + q at 0. Now F1=F2=Q24πϵ0a2 and F3=Q24πϵ0(2r)2=Q24πϵ0(2a2) and F′=14πϵ0.q(Q)r2=2qQ4πϵ0a2 The resultant of F1 and F2 is given by F=√F21+F22=√2F1=√2Q24πϵ0a2 The force F and F3 act along AP. Hence the net force acting on charge -Q at A due to charges -Q at B, C and D is F′′=F+F3 =√2Q24πϵ0a2+Q24πϵ0(2a2)=Q2(1+2√2)4πϵ0(2a2) For equilibrium, F' - F" = 0 or F'' = F', i.e., =Q2(1+2√2)4πϵ0(2a2)=2qQ4πϵ0a2 or q=Q4(1+2√2) Hence the correct choice is (a).

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