Four charges equal to -Q are placed...
Question

# Four charges equal to -Q are placed at the four corners of a square and a charge q is at i centre. If the system is in equilibrium the value of q is (2004 4)-(1+2V2) B (1+212) 0-2(1+212) D 2 (1+212)

NEET/Medical Exams
Physics
Solution
103
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e) [ a{4,11,19) ] ( Rightarrow f_{1}=F_{2}=frac{1}{4 pi epsilon_{0}} frac{theta^{2}}{a^{2}} ) ( Leftrightarrow quad Rightarrow quad F_{3}=frac{1}{4 pi epsilon_{0}} frac{(-Q)(-theta)}{(A+1)^{2}}=frac{1}{4 pi in(sqrt{2} a)^{2}} ) ( F N=sqrt{F_{1}^{2}+F_{2}^{2}}=sqrt{2} times frac{1}{4 pi C_{0} a^{2}} ) Sinu; Chary of ef are in Equilibrium ( F_{3}+F_{11}=f_{4} ) [ frac{1}{4 pi E_{0}left(2 a^{2}right)}+frac{sqrt{2} times Q^{2}}{4 pi E_{0} a^{2}}=frac{1}{4 pi epsilon} frac{Q(q)}{left(frac{a}{sqrt{2}}right)^{2}} ] ( frac{theta^{2}}{2 a^{2}}+frac{sqrt{2} x theta^{2}}{a^{2}}=frac{28 q}{a^{2}} ) si ( frac{a^{2}}{2}+sqrt{2} a^{2}=202 ) [ begin{array}{r} frac{0+0.52}{2}=22 Rightarrow q=frac{theta}{4}+frac{8}{sqrt{2}} 3 q=frac{theta}{4}[1+2 sqrt{2}] end{array} ]