what is the magnitude of the e field at the 3rd vertix?

Question

Two point charges of +10 nC and -20 nC are fixed at 2 vertices of an equil. triangle with sides of 2m, located in a vacuum. What is the magnitude of the E field at the third vertix.

I need help setting up this equation. i'm almost positive you use cos 60 and the answer is 7.8x10^4 N/C.

Answer 1

Let's do it slowly.

E=kq/r^2
k- constant
q point charge
r – distance from the charge to the point in question


We have to use a superposition principle using vectors, of course. In the equilateral triangle with two point charges at the base we will have


Total vertical Ev= E1 sin(60) + E2sin(120) since
sin(60)= sin(120)
We have
Ev=(E1+E2)sin(60)

Total horizontal or Eh= E1 cos(60) + E2cos(120) since
cos(60)=- cos(120) Eh=0
Eh= (E1-E2) cos(60)
So

Since
k= 8.988×109 N m^2 /C^2
r in both cases 2m

Ev=(k/r^2)(q1+q2)sin(60)
Ev=(8.988×109 /2^2)(-10e-9-20e-9)sin(60)
Ev=-19.5N-C (downward)

Now solving for Eh
Eh= (E1-E2) cos(60)
Eh=(k/r^2)(q1-q2)cos(60)
Eh=(8.988×109 /2^2)(10e-9+20e-9)cos(60)
Eh=33.7 N-C
E=sqrt(Eh^2+Ev^2)
E=38.9 N-C

I hope it helps.