Given an equilateral triangle with sides 20-cm, with point charges at each point:
q1 at the top with a charge of +4.0uM,
q2 at the bottom right with a charge of -4.0uM, and
q3 at the bottom left with a charge of +4.0uM.
k is assumed to be exactly 9.00x109 N-m2/c2
Question: What is the electric field at the center of the triangle?
E = (kq)/r2
The Attempt at a Solution
First and foremost, I employed a little geometry find the center of the triangle:
r = sqrt( (0.20-m)2 - (0.10-m)2 ) / 2
r = 0.866-m, rounded to
r = 0.90-m
E1 = (9.00x106 N-m2/c2)(4.0x10-6C) / (0.09-m)2
E1 = 4.4x106 N/C
The magnitudes of all three charges are equivalent, so:
E1 = E2 = E3
So the x and y components would be:
Ex = (4.4x106 N/C)cos30° + (4.4x106 N/C)cos330°
Ex = 7.6 x 106 N/C
Ey = (4.4x106 N/C)sin270° + (4.4x106 N/C)sin30° + (4.4x106 N/C)sin330°
Ey = -4.4x106 N/C
E = sqrt[ (7.6 x 106 N/C)2 - (4.4x106 N/C)2 ]
E = 6.2 x 106 N/C
So, my answer is 6.2 x 106 N, but the book answer key shows 5.4 x 106 N/C. That is quite a significant difference, but I can't seem to find where I'm making a mistake. It's driving me mad!