Question

point charges are placed at the vertices of a square of side a as calculate the magnitude of ratio q/Q so that net force on each Q is zero​

Answer 1

Answer:

(a) Electric force =

$\frac{ {kq}^{1} {q}^{2} }{ {r}^{2} }$

Electric force on Q by other charge Q=

$\frac{ {kQ}^{2} }{ {2a}^{2} }$

force on Q due to both q=

$\sqrt{2} \: \frac{kqQ}{ {a}^{2} }$

The resultant electric force =

$\frac{ {kQ</p><p>}^{2} }{ {2a}^{2} } \: + \sqrt{2} \: \frac{kqQ}{ {a}^{2} } </p><p>$

(b)Potential energy of two charges =

$\frac{ {kq}^{1} {q}^{2} }{r}$

potential energy of system =