A ray of light falls normally on a rectangular glass slab.
Draw a ray diagram showing the path of the ray till it emerges out of the slab.

Explanation

From fig `T cos theta = mg, T sin theta = F`
`(F)/(mg) = tan theta = (x)/(sqrt(1^(2)-x^(2)))` If `theta` is small `Tan theta = sin theta ~~ theta = (x)/(1), (F)/(mg) (x)/(1)`
Where `F= (1)/(4pi in_(0))(q^(2))/(4x^(2)) rArr (q^(2))/(16pi in_(0)x^(2) mg) = (x)/(1)`
`rArr x= ((q^(2)1)/(16pi in_(0)mg))^(1//3)`
`rArr` Tension in the string `T= sqrt(F^(2) + (mg)^(2))`
In the above case if the balls are suspended in a liquid of density `rho` and the distance between the balls remains the same, then
`(F)/(mg) =(F^(1))/(mg^(1)) rArr (F)/(F^(1)) = (mg)/(mg^(1)) rArr (mg)/(mg(1-(rho)/(d)))( because k = (F_(a))/(F_(m)))`
`rArr (1- (rho)/(d)) = (1)/(K) or K = (d)/((d-p))`
(d is density of material of the ball)
`rArr` In the above case if the charges on teh two balls are different and the angles made by those two strings to the vertical are `theta_(1) and theta_(2)` respectively, then `theta_(1) = theta_(2)` (as F, mg are same ) if `m_(1) gt m_(2) " then " theta_(1)lt theta_(2)`
`rArr` In the above application if the whole setup is kept in an artifical satellite (in zero gravity region) the angle between the two strings is `180^(@)` and tension in each string is `(1)/(4pi in_(0))(q^(2))/(41^(2))`