# Distrubution of Charges

1. Jun 29, 2008

### purduegirl

1. The problem statement, all variables and given/known data

Calculate the magnitude of the electric field at the origin due to the following distribution of charges: -q at (x,y) = (a,a), -q at (a,-a), +q at (-a,-a) and +q at (-a,a). Where q = 8.30 × 10-7 C and a = 5.15 cm.

2. Relevant equations

E = KQ/R^2 or KQ/2a^2
The summation of E = E1 + E2 + E3 + E4

3. The attempt at a solution

I calculated this with the + and - signs and got 0 because their are two signs for each one and they are all the same charge. I did this without the signs and got 5.63E6 N/C.

2. Jun 29, 2008

### Hootenanny

Staff Emeritus
Although the charges are of equal magnitude, they are not symmetrically distributed (i.e. there are two negative charges on the right of the square and two positive charges on the left); this results in a non-zero electric field anywhere in space.

3. Jul 17, 2008

### purduegirl

I knew there was an electric field here because with the orientation of the charges, when compared to a compass, would make the theoretically compass point in the east direction. But, from there, I have no idea how to solve this problem, since the method suggested by my prof. is not correct for my layout of the problem. The in class prob, had alternating charges.

4. Jul 17, 2008

### LowlyPion

Recall that if you have an electric field, you are dealing with a vector field.

In the case where you have alternating symmetrical and equal charges then at the mid point of the array of charges you have vectors all canceling out in both the x and y direction.

But as you have noted the charges are not alternating and hence you have a field at the origin. (They are symmetrical about the x-axis though.)

The way I would approach this problem then is to observe the geometry of the charges. Since they are on a square of 2*a on each side, you effectively have two charge pairs. These charge pairs are each separated by $$2 a \sqrt 2$$.

If you draw the vectors carefully you should see that the y components will indeed cancel out because of the x-axis symmetry. You should be able to answer the question then based on it being at the mid-point of a line separating 2 charges (each of which is now 2q) that are separated by $$2 a \sqrt 2$$.

Last edited: Jul 17, 2008