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Finding the field strength of an EF

  • Thread starter Watsonb2
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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:

[tex]\vec{E}[/tex]([tex]\vec{r}[/tex] = [k (Q) ([tex]\vec{r}[/tex] - [tex]\vec{R}[/tex]] / [([tex]\vec{r}[/tex] - [tex]\vec{R}[/tex])^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/[tex]\sqrt{2}[/tex], 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

  • #2
rl.bhat
Homework Helper
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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.
 

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