A particle a having a charge of `5.0 xx 10 ^(-7) C` is fixed in. a vertical wall. A second particle B of mass 100 g and. having equal charge is supended by a silk thread. of length 30 cm form the wall. The point of suspension is. 30 cm above the particle A. Find the angle of the thread. with the vertical when it stays in equilibrium.

Explanation

The situation shown in figure
Here the forces acting on bob B can be shown as
FBD of B is
Using Lami's theorem, we get
`(mg)/(sin((pi)/(2)+(theta)/(2)))=(F)/(sin(pi-theta))`

or `(mg)/(cos.(theta)/(2))=(Kq^(2))/(2xx0.30xxsin.(theta)/(2)xxsintheta)`
or `(mg)/(cos.(theta)/(2))=(Kq^(2))/(0.60sin.(theta)/(2)xx2sin.(theta)/(2)cos.(theta)/(2
))`
or `sin^(2).(theta)/(2)=(Kq^(2))/(2mg(0.60))`