Charges of +2 +2, and ` -2 muC ` respectively are placed at the vertices of an equilateral triangle of side ` 0.3m ` each in Find the net forces experienced by each charges.

Explanation

Here AB = AC =BC =0.3 and `q_A =q_B =+ 2mu C = 2xx 10 ^(-6) C and q_C =- 2muC =-2 xx 10 ^(-6)C `
Thus , it is clear that mutual force of attraction/repulsion between any two charges have equal magnitude given by
` " " F= |oversetto (F_(AB)) |=|oversetto (F_(CB)) |=(9xx 10 ^(9) xx (2xx 10 ^(-6))^(2)) /( (0.3)^(2)) =0.4 N `
Direction of these forces have been shown in
(a) As angle between `oversetto (F_(CA)) and oversetto (F_(CB)) is 60^(@) ` , hence
` |oversetto (F_C) | =2F _(CA) cos ""( (60)/( 2))^(@) `
` " " =2 xx0.4 xx (sqrt( 3))/(2)`
` = 0.69 N and ` the force is `TT `to AB.
(b) As angle between `oversetto (F_(AB)) and oversetto (F_(AC)) is 120 ^(@) ` , hence
` |oversetto (F_A) | 2F_(AB) cos 60^(@) = 2xx 0.4 xx (1)/(2) = 0.4 N `
and the force makes an angle ` 60^(@) ` from line AC.
(c) As in (b) force `|oversetto (F_B)| =0.4 N ` inclined at an angle of ` 60^(@) ` from line BC.
` (##U_LIK_SP_PHY_XII_C01_E10_022_S01.png" width="80%">