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 is shown in figure . Supppose the
piont suspension is O and let theta be the angle between
the thread and the vertical. Forces on the particle B are
(i) weight mg downward
(ii) tension T along the thread and
(iii) electric force of repulsion F along AB.
For equilibrium, these forces should add to zero. Let
X'' BX be the line perpendicular to OB. we shall take the
components of the forces along BX. This will giva a
relation between F, mg and `theta`
The verious andgles are shown in the figure. As
`OA= OB, /_OAB =` 90^@ - (theta)/(2)`
The other angles can be writen down directly.
Taking components along BX, we get
`F cos (theta) /(2) = mg cos ( 90^@ - theta)`
`= 2 mg sin (theta)/(2) cos (theta)/(2)`
or, `sin (theta)/(2) = (F)/ (2mg)`
Now,` F = (9xx10 ^9 Nm^2 C^(-2) xx (5.0 xx 10 ^(-7) C)^2 xx (1)/(AB^2)`
and `AB = 2 (OA) sin (theta)/(2).`
Thus, `F = (9 xx 10 ^9 xx 25 xx 10(-14))/(4 xx (30xx10^(-2))^2 xx sin^2. (theta)/(2)) N`
from (i) and (ii),
`sin (theta)/(2) = (F )/(2 mg) = (9xx 10^9 xx 25 xx 10 ^(-14) N )/(4xx ( 30 xx 10 ^(-2)^2 xx sin^2 .(theta)/(2)) . (1)/(2mg) or
`sin^3 (theta)/(2) =(9xx10^9 xx 25 xx 10 ^(-14) N)/(4xx9xx 10^(-2) xx 2 xx (100xx 10^(-3)kg) xx 9.8 ms(-2))`
=0.0032
or sin (theta)/(2) = 0.15, giving theta =17^@.`