# 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)]

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

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

**
**