# Homework Help: Electric field at the center of a square

1. Sep 25, 2008

### ally1h

1. The problem statement, all variables and given/known data
A charge is placed at each corner of a square. The charges at the top corners are each +6μC and the ones at the bottom are each -4μC. Each side of the square has a length of 10.0 cm. Determine the electric field strength at the center of the square.

2. Relevant equations
E = Kq/2a^2

3. The attempt at a solution
I understand that the electric field at the center is the vector sum of the fields due to each of the point charges. I somewhat understand the problem if all the charges were the same, then the electric field strength would be 0. But I'm a little thrown with multiple charges. Is it:

E = K(q1q2)/(2)(a^2)
E = (9x10^9 Nm^2/C^2)*[(6x10^-6 C)(4x10^-6 C) / (2)(.01 m^2)
E = 10.8 N/C

Somehow I don't think so because this doesn't really give me N/C as the answer. So... considering it is supposed to be the vector sum.. am I just supposed to add the charges together? In which case q = (6μC+6μC+ (-4μC)+ (-4μC)) = +4μC?

2. Sep 25, 2008

### Defennder

No you don't add the charges together. Note that the top corners have positive charges and the bottom ones have negative charges. Symmetry alone allows you to conclude that the field only has a y component. Just find the Ey contributions by the top and bottom charges. An easy way would be to find the Ey due to a single charge at a square corner at the top and separately for the bottom. Then just mutiply the answer by two to get the Ey contribution from the other half of the square.

3. Sep 25, 2008

### ally1h

Okay, let me see if I understand correctly. Since there is only a y component I should be finding the value of the vector??

Since the shape is a square the value of theta = 45 degrees. Since two sides are 5cm I should use the Pythagorean theorem to find the value of the vector?

I'm still confused. I'm very weak in geometry and trig

4. Sep 27, 2008

### Defennder

Yes you need to find the y component of the E-field. Draw a right-angle triangle with E_y and E_x as it sides and E as the hypotenuse. The angle between E_y and E is the same as the angle between the point charge on the square corner and the line from that corner charge to the centre. So we have E_y = E cos theta, if theta is the angle as described above.