A system consists of three charges at the vertices of an equilateral triangle...electric field?

Question

A system consists of three charges q1=5 x10^-6, q2=5 x 10^-6, q3= -5 x 10^-6 at the vertices of an equilateral triangle of sides d=2.75 cm. Find the magnitude of the electric field at a point halfway between the charges q1 and q2. Find the magnitude of the electric field halfway between the charges q2 and q3.

Ok, for the first question I know that the forces from q1 and q2 cancel out so all we have to worry about is q3. I think the correct equation is E= kq/r^2. I found r to be .0238 by using .01375^2 + b^2= .0275^2, but when I plug this into the equation, I am not getting the correct answer. What am I doing wrong?

Thank you!

Answer 1

To be clear, the question you've posted asks about fields, not forces--these two are not the same.  A resulting force is experienced by a charge within an electric field.

The electric field at any location is due to the vector summation of the three individual point charges' fields.  Each of the three charged particles contribute at every location in space.  Be sure you're mindful that these numbers also have direction and cannot simply be added; again, this is because electric field is a vector.

When i was still teaching, my first suggestion to any student was draw a diagram.  If you can correctly represent it, it's likely you've just improved your knowledge of the question and chances of success at the answer.

Answer 2

q1=5 x10^-6 C, q2=5 x 10^-6 C, q3= -5 x 10^-6 C

|qi|=q

d=2.75 cm

k~9*10^9 Nm^2/C^2

1. The magnitude of the electric field at a point

halfway between the charges q1 and q2:

You are correct: E= kq/r^2, where r=0.0238 m

r=(d*sqrt3)/2

E=(9*10^9)*5*10^-6/(0.0238)^2

E=79.4*10^6 V/m

2. The magnitude of the electric field at a point

halfway between the charges q1 and q2:

E=sqrt[(E2+E3)^2+E1^2], where

E2=E3=kq/(d/2)^2=4kq/d^2

and E1=kq/r^2=4kq/(3d^2)

E=(kq/d^2)sqrt(8+4/3)

E=(2kq/d^2)sqrt(7/3)

http://hyperphysics.phy-astr.gsu.edu/Hba...