Question

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}
]

# 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)

Solution