At each of the four corners of a square of side a, a charge +q is placed freely. What charge should be placed at the centre of the square so that the whole system be in equilibrium? [Ans: (-q/4)(1+2√2)]

for whole system to be in equilibrium the net force on each of  the charges must be zero and the net net force on the centre must be zero 
since the force on the charge which is placed on the corner side is the vector sum of the other three charges which are placed on the corner and  the chrge which is placed on the centre 
now let us consider a charge which is placed at corner of the squre of side a ,suppose their are 4 corner
so the net force on the charge which is placed at corner 1 will be 
F1 = F12 + F13 +F14 + Fc
where  F1 = net force on charge at corner 1 
F12 = force on 1 due to chage placed at corner 2 
F13 = force on charge 1 due to charge placed on corner 3 
Fc = force on 1 due to centre charge 

we know that force  F = k q1q2a^2 ,where k is constantF12 = q^2a^2(1,0)F14 = q^2a^2(0,1)F13 = q^22a^2(1,1)Fc = 2qQa^2 ( where r = a2)now the whole system is in equilibrium so F1 = F12+F14+F13+Fc         = q^2a^2(1,0) +q^2a^222(1,1)+q^2a^2(0,1) +2Qqa^2=0by taking q^2a^2 common to all and solving for Q in terms of q ,we get F1 = q^2a^2(1+122+Qq2)=0so Q2 q = -(1+122)Qq = -(12+14)now divide and multiply by4 in rhs ,we get Qq=- 44(12+14)Q = -q4(22+1)Q =-q4(1+22)   ans....


  • -10
What are you looking for?