Two small -sized identical equally charged spheres each having mass 1 mg are hanging in equilibrium as shown in the figure . The length of each string is 10 cm and the angle `theta` is `7^(@)` with the vertical . Calculate the magnitude of the charge in each sphere . (Take g =`10ms^(-2))`

Explanation

If the spheres are neutral the angle between them will be `0^(@)` when hanged vertically . Since they are positively charged spheres , there will be a repulsive force between them and they will be at equilibrium with each other at an angle of `7^(@)` with the vertiction. We can draw a free body diagram for one of the charged spheres and apply Newton.s second law for both vertical and horizontal directions . The free body diagram is shown. In the x -direction the acceleration of the charged sphere is zero . Using Newton.s second law `(vecF_(m)=mveca)` , we have T sin `thetahati-F_(e)hati=0`
T`sin theta=F_(e)`
Here T is the tension acting on the charge due to the string and `F_(e)` is the electrostatic force between the two charges . In the y -direction also the net acceleration experienced by the charge is zero . T `cos thetahatj-mg hatj=0`
Therefore T cos `theta =mg ` . By dividing equation (1) by equation (2) , `tan theta = (F_(e))/(mg)`
Since they are equally charged the magnitude of the electrostatic force is
`F_(e)= k""(q_(2))/(r^(2))` where k `=(1)/(4piepsilon_(0))`
Here r = 2a = 2L `sin theta`. By substituting these values in equation (3) ,
`tan theta = k(q^(2))/(mg(2L sin theta)^(2))`
Rearranging the equation (4) to get q
`q=2L sinthetasqrt(("mg tan" theta)/(k))=2xx0.1xxsin7^(@)xxsqrt(10^(-3)xx10xxtan7^(@))/(9xx10^(9))`
`q = 8.9xx10^(-9)C = 8.9nC`