# Finding the field strength of an EF

## Homework Statement

A square, with charges at each corner. The upper left (1) has a charge of +q. The upper right (2) has a charge of -2q. The bottom right (3) has a charge of +2q. The bottom left (4) has a charge of -q. The square has side length a.

a = 5.20 cm and q = 11.8 nC

Find the EF at the center of the square.

## Homework Equations

This was the equation that my prof gave us to deal with an observation point (in this case the center of the square) introduced into an Electric Field:

$$\vec{E}$$($$\vec{r}$$ = [k (Q) ($$\vec{r}$$ - $$\vec{R}$$] / [($$\vec{r}$$ - $$\vec{R}$$)^3]

Where r = the actual distance from the source charge to the observation point and R = the distance of the source charge from the origin.

## The Attempt at a Solution

I started by determining the distances between the source charges and the observation point, which all were a/$$\sqrt{2}$$, by pythagorean theorem.

I plugged this into the equation and got an answer of zero, which immediately flagged my attention, but I can't see why it came out that way. I then plugged the numbers into the equation for the second charge and got an answer of -1.57 x 10^14. This seemed very large, even though the charge of q is fairly large to begin with...

I left off there, maybe you guys could help me out and point out where I'm making my error...

Thanks,

-B

## Answers and Replies

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rl.bhat
Homework Helper
Electric fields due to charges 1 and 3 are in the opposite direction. The net field is towards charge 1. Similarly net field due charges due to 2 and 4 is towards charge 4. These two net fields are perpendicular to each other. Now find the resultant field due to all the four charges at the center.