Two path balls carrying equal charges are suspended from 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. `((1)/(sqrt(2)))^(2)`
B. `((r)/(.^(3)sqrt(2)))`
C. `((2r)/(sqrt(3)))`
D. `((2r)/(3))`

Answer

b

Explanation

Case-1:
`T sin theta= F_(e)` (i)

`T sin theta= F_(e )` (i)
`T cos theta= mg` (ii)
`tan theta= (F_(e ))/(mg)implies F_(e )= mg tan theta`
Here, `F_(e)= k(q^(2))/(r^(2))= mg .(r//2)/(y)` (iii)

`T_(1) sin theta_(1)= F_(e )` (iv)
`T_(1) cos theta_(1)= mg` (v)
`tan theta_(1)= (F'_(e ))/(mg)implies F'_(e )= mg tan theta_(1)` (v)
Here `F'_(e )= k(q^(2))/(r^(2))= mg.(r'//2)/(y//2)` (vi)
From (iii) &(vi)
`(kq^(2)//r^(2))/(kq^(2)//r'^(2))= (r)/(2r')`
`implies ((r'^(2))/(r))=(1)/(2)` or `r'=(r)/(2^(1//3))`