Electric Field at Origin: The Answer is 0

[don23]

Can someone help with this? I want to say the answer is zero but I don't know how to explain it.

Calculate the electric field at the origin due to the following distribution of charges: +q at (x,y)=(a,a), +q at (-a,a), -q at (-a,-a) and -q at (a,-a).

 

[whozum]

Each charge has an E field described by

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

Just find the vector sum at the origin, or better yet draw it out and see if you can cancel by symmetry.

 

[don23]

thanks

Thank you for the reply.
would i be correct is saying that

E= (+q*k/a^2+a^2)+(+q*k/a^2+a^2)+(-q*k/a^2+a^2)+(-q*k/a^2+a^2)= 0

??

 

[whozum]

Your conclusion is correct, but you measured the distance from the origin incorrectly, remember the distance between the origin and the point (x,y) is

$$\sqrt{x^2+y^2}$$

 

[don23]

thank you

thanks again.
in this case wouldn't the distance from the origin be (sq rt of (a^2+a^2))? In the original problem all the charges are at (a,a),(-a,a),(-a,-a) and (a,-a).

 

[whozum]

Yes that's correct, it simplifies to [itex] \sqrt{2}a[/tex]