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# Electric Field in Equilateral Triangle Watch

1. An equilateral triangle is made of 3 rods, each 0.1m in length. Rods AB and BC are charged +18nC and rod AC is charged -18nC. What is the electric field at the center of the triangle?

Here is my solution, but it is apparently incorrect:

Since all of the rods have the same magnitude of charge, they all generate the same magnitude of field.

r=1/2h where h=height of the triangle.
h=sqrt(L^2-(.5L)^2)=sqrt[(.2^2m)-(.1^2)]=0.173m
h/2=0.0866m

E=kq/r^2=(8.99x10^9)(18x10^-9)/(0.0866^2)
E=2.158x10^4N/C

In the triangle, EABx=-EBCx and EACx=0, so we only need to focus on the y direction.

Since this is an equilateral triangle, all of the angles are 60 degrees

My thought would be that Enet=EABY+EBCY+EAC
so Enet=2Esin(60)+E=2(2.158x10^4sin (60))+2.158x10^4
Enet=5.89x104N/C
This is the answer I got, but apparently it is incorrect. Any idea where I went wrong?
2. (Original post by PatchworkTeapot)
An equilateral triangle is made of 3 rods, each 0.1m in length. Rods AB and BC are charged +18nC and rod AC is charged -18nC. What is the electric field at the center of the triangle?

Here is my solution, but it is apparently incorrect:

Since all of the rods have the same magnitude of charge, they all generate the same magnitude of field.

r=1/2h where h=height of the triangle.
h=sqrt(L^2-(.5L)^2)=sqrt[(.2^2m)-(.1^2)]=0.173m
h/2=0.0866m

E=kq/r^2=(8.99x10^9)(18x10^-9)/(0.0866^2)
E=2.158x10^4N/C

In the triangle, EABx=-EBCx and EACx=0, so we only need to focus on the y direction.

Since this is an equilateral triangle, all of the angles are 60 degrees

My thought would be that Enet=EABY+EBCY+EAC
so Enet=2Esin(60)+E=2(2.158x10^4sin (60))+2.158x10^4
Enet=5.89x104N/C
This is the answer I got, but apparently it is incorrect. Any idea where I went wrong?
Have you been told to use E=kq/r^2? That's not the electric field from a rod...

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Updated: September 18, 2016
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