- #1

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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).

- Thread starter don23
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- #1

- 10

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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).

- #2

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[tex] \vec{E} = \frac{kq}{\vec{r}^2} [/tex].

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

- #3

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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

??

- #4

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[tex] \sqrt{x^2+y^2} [/tex]

- #5

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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).

- #6

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

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