4 Charges at the corners of a square

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Homework Statement



Consider a square which is 1m on a side. Charges are placed at the corners of the square as follows:

+4 uC at (0,0)
+4 uC at (1,1)
+3 uC at (1,0)
-3 uC at (0,1)

What is the magnitude of the electric field at the square's center?

Homework Equations



E= k(Q/r^2)

(sorry i suck at latex, don't know how to use it)

The Attempt at a Solution



Well I thought this was fairly simple. I just plugged in Q and for r^2 i used cos(45)^2 since the the distance to the center is 1cos(45) = sqrt(2)/2

This gave me 3 different values. for the 4 uC's i got 71934.4 N/C, for the +3 i got 53950.8 and for the -3 i got -53950.8.

So these are the fields in their respective direction. To calculate the magnitude I sum them all, square them and then take the square root. That gives me 143868. However this is not one of the answers:

1.1E5
1.3E5
1.5E5
1.7E5

Anyone have any idea if i did something wrong?
 
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Could you show please the direction of the individual field vectors?

ehild
 
[PLAIN]http://img831.imageshack.us/img831/1600/unledss.jpg

I hope this is right

1 2 and 4 are positive whereas 3 is negative
 
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It is right. What can you say about the field of the two 4μC charges? What is the direction of the resultant?

ehild
 
Well it should cancel right? Since they are the same magnitude and opposite direction. So would i just sum field's 3 and 2, square them and then take the square root?
 
Well, you add only those two fields. You know their magnitudes, don't you? and they point in the same direction, so what is the magnitude of their sum? Why do you square and take the square root?

ehild
 
The reason why i square and take the square root is because my book says so :P But i realized that I already know the magnitude and direction (in the book it was calculated separately) so I just need to add them which gives me 107900. Is that correct? Would the answer be 1.1E5?
 
Ortix said:
The reason why i square and take the square root is because my book says so :P But i realized that I already know the magnitude and direction (in the book it was calculated separately) so I just need to add them which gives me 107900. Is that correct? Would the answer be 1.1E5?


Yes. :smile:

ehild