Electric Field of Point Charges in Equilateral Triangle

[END]

Homework Statement

A point charge q =+6.4 \mu \mathrm{C} is placed at each corner of an equilateral triangle with sides 0.29 \mathrm{m} in length.
What is the magnitude of the electric field at the midpoint of any of the three sides of the triangle?

Hint: if you are careful to examine the symmetry of this problem before starting to calculate, you will notice that the problem is much shorter than you think!

Homework Equations

$$E=\frac{kq}{r^2}$$

The Attempt at a Solution

[/B]
I suspect that the hint alludes to the fact that at the midpoint of any given side, the electric field of the two adjacent charges whose position lie along the same axis as the midpoint will in effect cancel each other out. Therefore, the only electric field I need to calculate is that of the point charge furthest from the midpoint of that given side.

I know the charge q =+6.4\times 10^{-6} \mathrm{C} and that the distance r between the point charge and midpoint is the height of the equilateral triangle so that r=\frac{\sqrt{3}}{2}\times.29\mathrm{m}.

If I plug these into the equation, I get
$$E=k\frac{6.4\times 10^{-6} \mathrm{C}}{(\frac{\sqrt{3}}{2}\times.29\mathrm{m})^2}$$
$$=
\boxed{911935.4964~ \mathrm{N/C}}$$

Is this the correct answer?

Thank you.

 

[Simon Bridge]

The reasoning is sound ... havn't checked the arithmetic.

 

[END]

Thank you for your feedback, Simon! The answer I posted was indeed correct.

 

[Simon Bridge]

Well done.

Just a note:
I may have said that the answer was right or wrong fersure, but how do you know I got it right?
You are training to work on problems where nobody knows the right answer, so there is nobody to ask.
As you progress you need to start thinking about how you can tell that you have it right without having to ask someone.