Question

Charges of +2microC, +3microC and -8microC are placed in air at the vertices of an equilateral triangle of ide 10cm.What is the magnitude of the force acting on the -8microC due to the other two charges?

Answer 1

Let charge `2 muC,3muC,-8 muC` are placed at point `A,B,C` of the triangle shown.

So,net force on `-8 muC` due to `2muC` will act along `CA`

and the value is `F_1=(9*10^9*(2*10^-6)*(-8)*10^-6)/(10/100)^2=-14.4N`

And due to `3muC` it will be along `CB` i.e `F_2=(9*10^9*(3*10^-6)(-8)*10^-6)/(10/100)^2=-21.6N`

So,two forces of `F_1` and `F_2` are acting on the charge `-8muC` with an angle of `60^@` in between,so the nect force will be, `F= sqrt(F_1^2 +F_2^2 + 2F_1 F_2 cos 60)=31.37N`

Making an angle of `tan^-1((14.4 sin 60)/(21.6+14.4 cos 60))=29.4^@` with `F_2`