Answer & explanation
Correct answer: option 3
$\text {Resultant Force on third charge will be } F_R = 2Fsin\theta = \frac{2kqQ }[{x^2 +\frac{d^2}{4}}]^3/2$
$\text{For $F_R$to be maximum } x = \frac{d}{2\sqrt 2}$
Correct answer: option 3
$\text {Resultant Force on third charge will be } F_R = 2Fsin\theta = \frac{2kqQ }[{x^2 +\frac{d^2}{4}}]^3/2$
$\text{For $F_R$to be maximum } x = \frac{d}{2\sqrt 2}$