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°

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.

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