A point charge is placed at each corner of a square with side length $a$. The charges all have the same magnitude $q$. Two of the charges are positive and two are negative, as shown in the following figure.

What is the magnitude of the net electric field at the center of the square due to the four charges in terms of $q$ and $a$?

I'm trying to understand this question by using the solution my instructor gave me, but I'm confused.

I found $r=\frac{a}{\sqrt{2}}$, which makes sense. However then it says:

The magnitude of the electric field due to each charge is the same and equal to

$$E_q = \frac{kq}{r^2} = \frac{2kq}{a^2}$$

How did they come up with this? I only know the equations $F = \frac{k\lvert q_1q_2\rvert}{r^2}$ and $E = \frac{F_0}{q_0}$ and I can't seem to work out how they got to this point.

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    $\begingroup$ I just wanted to say, I consider this an excellent example of how to ask a homework question :-) $\endgroup$ – David Z Sep 26 '12 at 0:00
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    $\begingroup$ @DavidZaslavsky It could use the referenced diagram, though, since there are two distinct ways to put the charges on a square resulting in two different answers to the original question. $\endgroup$ – Mark Eichenlaub Sep 26 '12 at 1:10

Everything you have is correct. It's just an algebra step to get to the final answer.

You already have $$r = \frac{a}{\sqrt{2}}$$ Squaring it, you get $$r^2 = \frac{a^2}{2}$$ Now use this to make a substitution for $r^2$ in your equation for the electric field strength $$\frac{k q}{r^2} = \frac{kq}{a^2/2} = \frac{2 k q}{a^2}$$

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    $\begingroup$ Now that I see it, I get it. Perfect answer! $\endgroup$ – Cody Sep 26 '12 at 21:32

The equation comes if you put q1=q0 in the equations you have mentioned.


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