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## Homework Statement

Given an equilateral triangle with sides 20-cm, with point charges at each point:

*q*at the top with a charge of +4.0

_{1}*u*M,

*q*at the bottom right with a charge of -4.0

_{2}*u*M, and

*q*at the bottom left with a charge of +4.0

_{3}*u*M.

*k*is assumed to be exactly 9.00x10

^{9}N-m

^{2}/c

^{2}

Question: What is the electric field at the center of the triangle?

## Homework Equations

E = (

*kq*)/

*r*

^{2}

## 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

E

_{1}= (9.00x10

^{6}N-m

^{2}/c

^{2})(4.0x10

^{-6}C) / (0.09-m)

^{2}

E

_{1}= 4.4x10

^{6}N/C

The magnitudes of all three charges are equivalent, so:

E

_{1}= E

_{2}= E

_{3}

So the x and y components would be:

E

_{x}= (4.4x10

^{6}N/C)cos30° + (4.4x10

^{6}N/C)cos330°

E

_{x}= 7.6 x 10

^{6}N/C

E

_{y}= (4.4x10

^{6}N/C)sin270° + (4.4x10

^{6}N/C)sin30° + (4.4x10

^{6}N/C)sin330°

E

_{y}= -4.4x10

^{6}N/C

E = sqrt[ (7.6 x 10

^{6}N/C)

^{2}- (4.4x10

^{6}N/C)

^{2}]

E = 6.2 x 10

^{6}N/C

So, my answer is 6.2 x 10

^{6}N, but the book answer key shows 5.4 x 10

^{6}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!