Four charges equal in magnitude form a square

AI Thread Summary
Four equal charges of 20.0 microC are positioned at the corners of a square with a side length of 0.180m, and the task is to determine the electric field at the center. The distance from each charge to the center is calculated to be 0.127m, not 0.180m. Electric fields are vector quantities, so they must be added with both magnitude and direction considered, rather than simply summing their magnitudes. Charges 2 and 3 will cancel each other out due to their opposite nature, leaving the contributions from charges 1 and 4 to be combined. The direction of the electric field is determined by the nature of the charges, pointing towards negative charges and away from positive ones.
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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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