Question

Three charges with q = +6.3

Answer 1

q = charge = 6.3 x 10^-6 C

L = 15 cm = 0.15 m

in triangle AOC ::

AC = sqrt (AO² + CO²)

AC = sqrt (0.15² + 0.15²)

AC = 0.212 m

similarly , BC = 0.212 m

the force between two charges is given as ::

F = k q₁ q₂ /r²

force by A on C :

F_ca = k q_a q_c /AC²

F_ca = (9 x 10⁹) (6.3 x 10^-6 ) (6.3 x 10^-6 ) / (0.212)²

F_ca = 7.95 N

force by B on C :

F_cb = k q_b q_c /BC²

F_ba = (9 x 10⁹) (6.3 x 10^-6 ) (6.3 x 10^-6 ) / (0.212)²

F_ba = 7.95 N

each force F_ca and F_bc makes angle 30 with the y-axis

component of F_ca along X- and y- direction are given as :;

F_cax = F_ca Sin30 = 7.95 Sin30 = 3.975 N       towards positive X-axis

F_cay = F_ca Cos30 = 7.95 Cos30 = 6.885 N    towards negative Y-axis

component of F_cb along X- and y- direction are given as :;

F_cbx = F_cb Sin30 = 7.95 Sin30 = 3.975 N      towards negative X-axis

F_cby = F_cb Cos30 = 7.95 Cos30 = 6.885 N towards negative Y-axis

F_cax and F_cbx   are equal and opposite , hence these components cancel out

F_cay and F_cby are in same direction and they add together to give the net force

so F_net = F_cay + F_cby

F_net = 6.885 + 6.885

F_net = 13.77 N

since the net force in in negative Y-direction , counterclockwise it makes 270 degree angle with +X axis