Electrical field of an equialateral triangle

AI Thread Summary
An equilateral triangle is formed with three rods, each 10 cm long, where two rods carry a charge of +63 nC and one carries -63 nC. The electric field at the center of the triangle is calculated using the formula E_Rod = KQ / (d*sqrt(d^2 + (L/2)^2)), with K as the electrostatic constant. The distance from the center to each rod is 4.33 cm. Initial calculations yielded an incorrect net electric field due to a miscalculation of the triangle's height and the need to consider horizontal and vertical components of the forces. After correcting these errors, the problem was resolved successfully.
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Homework Statement


An equilateral triangle is formed with 3 rods, each with a length of 10cm. Two of them carry a charge of +63nC and 1 carries a charge of -63nC. The charge is uniformly distributed along each rod. What is the magnitude of the electric field at the center of the triangle?


Homework Equations


E_Rod = KQ / (d*sqrt(d^2 * (L/2)^2))


The Attempt at a Solution


Center of triangle is 4.33cm, or .0433m from each rod.

Constant K = 9.0*10^9
Q: Charge
d: Distance
L: Length of rod
E_Rod = KQ / (d*sqrt(d^2 * (L/2)^2))
E_Rod = (9.0*10^9)Q / (.0433sqrt(.0433^2 + .05^2))
Plugging in 63*10^9, 63*10^9, and -63*10^9, I receive:
E_Rod(1,2,3) = 197976,197976,-197976

Adding the three gives me a net E_rod = 197976, which is wrong.
 
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draw a picture of the forces involved. Split them up in horizontal and vertical components and add those.
 
Thanks.. Figured it out. Also figured out I was calculating the triangle height wrong.
 
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