Question

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)]

Answer 1

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

$$\begin{gathered} weknowthatforceF=k\frac{q1q2}{a^2},wherekiscons\tan t \\ F12=\frac{q^2}{a^2}(1,0) \\ F14=\frac{q^2}{a^2}(0,1) \\ F13=\frac{q^2}{2a^2}(1,1) \\ Fc=\frac{2qQ}{a^2}(wherer=\frac{a}{\sqrt[]{2}}) \\ nowthewholesystemisinequilibriumso \\ F1=F12+F14+F13+Fc \\ =\frac{q^2}{a^2}(1,0)+\frac{q^2}{a^22\sqrt[]{2}}(1,1)+\frac{q^2}{a^2}(0,1)+\frac{2Qq}{a^2}=0 \\ bytaking\frac{q^2}{a^2}commontoallandsolvingforQintermsofq,weget \\ F1=\frac{q^2}{a^2}(1+\frac{1}{2\sqrt[]{2}}+\frac{Q}{q\sqrt[]{2}})=0 \\ so \\ \frac{Q\sqrt[]{2}}{q}=-(1+\frac{1}{2\sqrt[]{2}}) \\ \frac{Q}{q}=-(\frac{1}{\sqrt[]{2}}+\frac{1}{4}) \\ nowdivideandmultiplyby4inrhs,weget \\ \frac{Q}{q}=-\frac{4}{4}(\frac{1}{\sqrt[]{2}}+\frac{1}{4}) \\ Q=-\frac{q}{4}(2\sqrt[]{2}+1) \\ Q=-\frac{q}{4}(1+2\sqrt[]{2})ans.... \end{gathered}$$



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