Three positive charge 3 esu, 4 esu and 5 esu are placed at the vertices of an equilateral triangle of side 10 cm. What is the force acting the largest charge?

Explanation

Three charges are placed at the corners of the equilateral triangle ABC. The largest charge i.e. 5 esu is situated at the point C.

Force acting on the charge at C due to the charge of A
`F_(1)=(3xx5)/((10)^(2))=0.15` dyn, along `vec(AC)`
Force acting on the charge at C due to the charge at B,
`F_(2)=(4xx5)/((10)^(2))=0.2` dyn, along `vec(BC0`
`:.` Resultant force acting on the charge at C,
`F=sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)cos 60^(@))`
`=sqrt((0.15)^(2)+(0.2)^(2)+2xx0.15xx0.2xx0.5)`
`=sqrt90.0925)=0.304`dyn