Four charges equal in magnitude form a square

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Homework Help Overview

The problem involves four charges of equal magnitude placed at the corners of a square, with the objective of determining the electric field at the center of the square. The charges are specified as 20.0 microC, and the side length of the square is 0.180 m.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the electric field by summing the contributions from each charge, but expresses uncertainty about the correctness of this method. Some participants question the distance used in the calculations and emphasize the need for vector addition of electric fields. Others explore the implications of charge interactions and the direction of the electric field vectors.

Discussion Status

Participants are actively discussing the calculations and the vector nature of electric fields. Some have provided clarifications regarding the distance to the center and the method of vector addition. There is an ongoing exploration of how the electric fields from different charges interact, particularly regarding cancellation and direction.

Contextual Notes

There are indications that the original poster may have miscalculated the distance from the charges to the center of the square. Additionally, the discussion includes considerations about the assumptions made regarding the addition of electric field magnitudes versus vectors.

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Homework Statement



Four charges equal in magnitude of 20.0 microC are placed on the four corners of a square with side length 0.180m. Determine the electric field at the centre of the square.

(-q) ---------- (+q)
l l
l l
l l
(+q)----------(+q)

1 --------------- 2

3-----------------4

Formatting messes up the square but you get the point.

Homework Equations



ɛ = kq / r
ɛ = delta V / r

The Attempt at a Solution



I tried just adding up the combinations of each charge after calculating ɛ = kq / r for each. I didn't feel confident this was the correct way to do it, but no answer or solution was given.

ɛ1 = k(-0.00002) / 0.180 = -1.00 x 10^6
ɛ2&3&4 = k(0.00002) / 0.180 = 1.00 x 10^6

ɛtotal = ɛ1 + ɛ2 + ɛ3 +ɛ4
= -1.0 x 10^6 + 3(1.0 x 10^6)
= 2.0 x 10^6
 
Last edited:
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Please show your work in more detail. There is no way we can tell if you did it right or wrong based on your post.
 
Orodruin said:
Please show your work in more detail. There is no way we can tell if you did it right or wrong based on your post.
Updated my solution to show my full work.
 
First of all, the distance from the charges to the center is not 0.18 m.

Second, the electric field is a vector. You cannot simply add the magnitudes of the fields from each charge. You need to add the field vectors as a vector addition, i.e., using both magnitude and direction.
 
Orodruin said:
First of all, the distance from the charges to the center is not 0.18 m.

Second, the electric field is a vector. You cannot simply add the magnitudes of the fields from each charge. You need to add the field vectors as a vector addition, i.e., using both magnitude and direction.

Right.

So I found the length to the centre from each corner to be 0.127m.

So if I were to add them as vectors would charge 2 and 3 cancel each other out, and I would be left to add charges 1 and 4 together since they are opposite charges?
 
Correct.
 
Orodruin said:
Correct.

Awesome thanks.

One last thing, is there a rule for determining which way the electric field vector is going? Like will it be towards 1 (-q) or 4 (+q)?
 
The field is pointing in the direction in which the force on a positive test charge at the given position would point. Thus, the field from a negative charge points towards it (the negative charge) and the field of a positive charge away from it (the positive charge).
 

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