# Four charges equal in magnitude form a square

• abm77
In summary: So in this case, the field at the center of the square will point towards charge 1 (-q) and away from charge 4 (+q).
abm77

## 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.

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:
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).

## 1. How do the charges arrange themselves in a square?

The charges will arrange themselves at the corners of a square, with each charge being equidistant from the other charges.

## 2. What is the force between two opposite charges in this arrangement?

The force between two opposite charges in this arrangement is attractive, as opposite charges attract each other.

## 3. Is there a difference in the force between two adjacent charges compared to two opposite charges?

Yes, the force between two adjacent charges is repulsive, as like charges repel each other. This is different from the attractive force between two opposite charges.

## 4. How does the magnitude of the charges affect the force between them?

The magnitude of the charges directly affects the force between them. As the magnitude increases, so does the force between the charges.

## 5. Can this arrangement of charges be used to demonstrate Coulomb's Law?

Yes, this arrangement is a common example used to demonstrate Coulomb's Law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

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