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January 30, 2015

January 30, 2015

Posted by **Jerry** on Sunday, October 17, 2010 at 10:02am.

- Physics -
**bobpursley**, Sunday, October 17, 2010 at 11:40amTHere are a number of ways to do this, some more difficult than others.

If you note the E due to q2 is upward, and the E due to q1 is horizontal, then the E due to q3 must add to zero. That is, the E due to q3 in the horizontal direction must be equal and opposite to q1, and The E due to q3 in the vertical must be equal and opposite to Q2

So dealing with q3 first.

letting o be the distance of the horizontal (o^2=d^2+4d^2=5d^2; or o=dsqrt5)

Ehorizontal=k q/o^2*2d/o=2kqd/o^3

Evertical= kq/o^2*d/o=kqd/o^3

So now setting the vertical of q3 equal to the vertical of q2

k*q2/d^2=kqd/o^3

q2=q (d/o)^3=q(d/dsqrt5)^3=q/5sqrt5

now, setting the horizontal of q3 equal to the horizontal of q1

kq1/4d^2=2kqd/o^3

q1=8q (d/o)^3 and you can finish the algebra.

Check all this math, I did it in my head keying it in.

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