A particle of mass m and charge q is suspended from a point A with a silk thread of length l. Another particle of charge `-q` is fixes at a distance I from the point A such that point A and the charge -q lie in same horizontal plane. If at equilibrium , the thread makes angle `30^(@)` with vertical, find the mass m and tension in the thread . ( Take acceleration due to gravity as g).

Explanation

The situation is shown in the figure . The triangle shows is equilateral.

`implies F+ ( 1)/( 4pi epsilon_(0))(q^(2))/(l^(2))`
Consider the free body diagram of charge q

For equilibrium in horizontal direction, `T sin 30^(@) = F sin 30^(@) implies T = F`
For equilibrium in vertical direction.
`F cos 30^(@) + T cos 30^(@) = mg `
`implies 2F cos 30^(@) = mg ` `[ :' T= F]`
`implies m = 2 xx ( 1)?( 4 pi epsilon_(0))xx(q^(2))/(l^(2))(cos 30^(@))/( g) = (sqrt(3)q^(2))/((4 pi epsilon_(0))l^(2) g)`
Also , `T = E implies T = ( 1)/(4pi epsilon_(0))(q^(2))/(l^(2))`