Two path balls carrying eqaul chareges are suspended froom a common point by strings of equal length, the strings are rightly clamped at half the height. The equilibrium separation between the balls, now becomes :

Options

A. `((2r)/(3))`
B. `((1)/(sqrt(2)))^(2)`
C. `(r/(root(3)2))`
D. `((2r)/(sqrt(3)))`

Answer

C

Explanation

If `theta` is angle which each pith ball in equibrium,makes with the vertical, then as is clear from Fig.

`tan theta = (r//2)/(y) = (F)/(mg) = (kq^(2)//r^(2))/(mg)` or `y = (mg r^(3))/(2 kq^(2))`
`:. (r')/(r ) = ((y')/(y))^(1//3) = ((y//2)/(y))^(1//3) = ((1)/(2))^(1//3)`
`r' = (r )/(2^(1//3))`