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Homework Statement
Three 18cm long rods form an equilateral triangle. Two of the rods are charged to +10 nC, and the third to  10 nC.
What is the electric field strength at the center of the triangle?
Homework Equations
$$ \vec{E} = \frac{k*q}{r^2} $$
The Attempt at a Solution
1. Draw the thing.
2. Integrate.
$$\int_0^d\frac{\left(k\cdot\frac{Q}{d}\right)}{\sqrt{\left(\left\frac{d}{2}x\right\right)^2+\left(d\left(\frac{d}{2}\right)^2\right)}}\cdot\left(\frac{\left(\left\frac{d}{2}x\right\right)}{\sqrt{\left(\left\frac{d}{2}x\right\right)^2+d\left(\frac{d}{2}\right)^2}}\right)dx$$
*d is in meters instead of centimeters
3. Add the vectors.
Because the charges are equal, and it is an equilateral triangle, I'm just going to simplify things.
The Y component is the bottom bar which is charge 10 with the magnitude of the integral above. The other two cancel.
The X component is 2*the integral of above from both the negative plate on the right and the positive plate on the left.
The magnitude is $$\sqrt{y^{2} + x^{2}}$$
For this problem the answer I computed is 0.0105401. I messed up somewhere and cannot find it...
For a sanity check, I know it must be less than the point charges at the center of the rods ~ sqrt(1.4^2+1.4cos(45)^2), but my answer seems too small.
Thanks in advance.
SOLUTION:
If I change the length of the height to sqrt((d^2(d/2)^2)/2 The integral does work: (Fixed)
$$\int_0^d\frac{\left(k\cdot\frac{Q}{d}\right)}{\left(\sqrt{\left(\left\frac{d}{2}x\right\right)^2+\left(\frac{\sqrt{d^2\left(\frac{d}{2}\right)^2}}{2}\right)^2}\right)}\cdot\frac{\left(\left\frac{d}{2}x\right\right)}{\left(\sqrt{\left(\left\frac{d}{2}x\right\right)^2+\left(\frac{\sqrt{d^2\left(\frac{d}{2}\right)^2}}{2}\right)}\right)}dx$$
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