Two small identical balls, each of mass 0.20 g, carry identical charges and are suspended by two threads of equal lengths. The balls position themselves at equilibirum such that the angle between the threads is `60^(@)`. If the distance between the balls is 0.5 m, find the charge on each ball.

Explanation

Given, `m=0.20 g =0.2xx10^(-3)` kg, `theta=60^(@) implies alpha =theta/2=30^(@)`
`r=0.5`m, Let `q_(1)=q_(2)=q`
From fig, `T sin alpha =1/(4pi epsi_(0))q^(2)/r^(2)` ...(1)
`T cos alpha = mg` ...(2)
`((1))/((2)) implies tan alpha =q^(2)/(4pi epsi_(0) r^(2) mg)`
`implies q^(2)=4 pi epsi_(0) r^(2) mg tan alpha`
`implies q^(2)=4 pi epsi_(0) r^(2) mg tan alpha`
`=1/(9xx10^(9))xx(0.5)^(2)xx0.2xx10^(-3)xx10xxtan 30^(@)`
`=1/(9xx10^(9))xx25xx10^(-2)xx0.2xx10^(-3)xx10xx1/sqrt(3)`
`=50/(9xxsqrt(3))xx10^(-14)`
`:.` Charge on each ball, `q=1.79xx10^(-7) C`.