Calculate the Electric Field at One Corner of a Square

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Homework Help Overview

The problem involves calculating the electric field at a corner of a square where three corners contain point charges. The square has a side length of 2.0 cm, and the charges are specified with their magnitudes and positions. The goal is to determine the electric field at the corner without a charge.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the electric field contributions from each charge and how to combine them as vectors. There is an exploration of the distances involved and the directions of the electric fields due to each charge.

Discussion Status

Participants are actively engaged in verifying their calculations and questioning the correctness of the provided answer. Some suggest that the answer key may contain an error, while others are attempting to reconcile their results with the expected outcome.

Contextual Notes

There is mention of different values for the constant K used in calculations, as well as confusion regarding the treatment of the negative charges in the electric field calculations. Participants are also addressing potential discrepancies in the expected answer.

jumbogala
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Homework Statement


One side of a square has length 2.0 cm.

In three of the square's corners, there are point charges.
Top left corner: -1.0 nC (call this A)
Bottom left corner: 10 nC (call this B)
Bottom right corner: -1.0 nC (call this C)

The top right corner has no charge, and is labelled P.

What is the magnitude of the electric field at point P?


Homework Equations


Electric Field = KQ/r2

The Attempt at a Solution


Electric Field Due to A:
K(-1.0 x 10-9 C)/(0.02 m2) = -22475 N/C, left

Electric Field Due to C:
Same as above, but this time the direction is down

Electric Field Due to B:
First calculate the distance from B to P. This is 0.022+0.022 = 0.0008. Take the square root of 0.0008 to find 0.0282 m.
K(10 x 10-9 C) / (0.0282)2 = 112375 N/C, at a 45 degree angle pointing north east.

Now, take the sum of all three parts to find the net electric field.
A and C together point south west, with a magnitude of 31784.44 N/C.

112375 N/C - 31784.44 N/C = 80590.55 N/C, north east.

However, the answer given on the page is not what I got. What went wrong?
 
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Remember they add like vectors.
 
I thought I did add them like vectors.

Adding vectors A (points left) and C (points down) gives a new vector which points south west...

22475^2 + 22475^2 = 1010251250.. take the square root of this and get 31784.44.

Since this vector points the opposite direction as vector B, which is 112375, can't I go 112375 - 31784.44?
 
So you have. Sorry I just calculated it and I get (5 -√2)*22,500 = 80,680

using k = 9*109

So I can't say what they are looking for then.
 
I get something similar, but I used K = 8.99x10^9. They are definitely looking for the electric field, but the answer given is 1.1 x 10 ^5.

I think whoever wrote the answer sheet probably made a mistake.

Thanks!
 
jumbogala said:
I get something similar, but I used K = 8.99x10^9. They are definitely looking for the electric field, but the answer given is 1.1 x 10 ^5.

I think whoever wrote the answer sheet probably made a mistake.

Thanks!

Yes their mistake. They canceled out the -1 nC charges. They don't as described.
 

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