Three identical point charges in an equilateral triangle.?

Determine the directions of the fields at the lower left corner
From the charge at the top, the field is at 240
From the charge at the lower right, the field is at 180
Both fields have the same magnitude E = k*3e-6/0.1² = 2,696,566 N/C
Find horizontal components
Ecos240 + Ecos180 = -4,044,398 = Ex
Find vertical components
Esin240 + Esin180 = -2,335,034 = Ey
Resultant = √Ex²+Ey²) = En = 4,670,069 at Arctan(Ey/Ex) = 210°
F = q*En = 14.01N at 210°...Show more

Get the direction by symmetry.

Then add the components of the two forces (acting on the lower left charge) in this direction.

The method is shown in the link....Show more

Set the lower left point at the origin.

q is the point at the origin.
q1 is the lower right point on the x-axis.
direction from q1 to q = e1_hat = -1i + 0j
q2 is the point at the top
direction from q2 to q = e2_hat = cos(240)i + sin(240)j

Definition of Electric field
E = kq/r^2

|E1| = k*q1/r^2 = 9*10^9 N*m^2/C^2 * 3*10^-6 C / (0.1 m)^2 = 2.7 * 10^6 N/C
|E2| = k*q2/r^2 = same thing = 2.7*10^6 N/C

E1 = |E1|*e1_hat = -2.7*10^6 i + 0 j N/C
E2 = |E2|*e2_hat = -1.35*10^6i - 2.34*10^6j N/C

Definition of Electrical Force
F = E*q

F1 = E1*q = 3*10^-6 C * (-2.7*10^6 i + 0 j) N/C = -8.1 i + 0 j N
F2 = E2*q = 3*10^-6 C * (-1.35*10^6i - 2.34*10^6j) N/C = -4.05 i - 7.02 j N

Resultant force on q = F1 + F2 = -12.05 i - 7.02 j N

magnitude of F = sqrt(i^2 + j^2) = sqrt((-12.05)^2 + (-7.02)^2) = 13.9 N

direction = atan(j/i) = atan(-7.02 / -12.05) = 210 deg...Show more