# Electric Field Corner homework problem

1. Mar 24, 2007

### kitty9035

1. The problem statement, all variables and given/known data
Positive charges are situated at three corners of a rectangle, as shown in the attachment. Find the electric field at the fourth corner.

the numbers are: bottom left- 3.00 nC bottom right-5.00 nC top right- 6.00 nC
the length of the rectangle is .6 m and the width is .2 m
2. Relevant equations

F=K q1/r^2

3. The attempt at a solution

I found all of the forces

F1= 9*10^9 (3*10^-9/.2^2) = 675 n/C

F2= 9*10^9(5*10^-9/.63^2) = 113 n/C

F3= 9*10^9(6*10^-9/.6^2) = 150 n/C

Then I know you have to combine vectors and get the answer, but that's what I'm having trouble doing.

#### Attached Files:

• ###### physics problem.bmp
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2. Mar 24, 2007

### robb_

Just a quick note, usually E is reserved for the field and F is for the force.
It is unclear to me as to whether the charges are positive or negative and this will determine the direction of the field. (your calculation seems to imply that they are all positive)
So once you have the magnitude of the individual components of the field, I would draw them on a vector diagram. Choose a coordinate system and break up each vector into components along each of the coordinate axes. Add like components to get the resultant field in components. Then you may express the final result in a variety of ways- leave it in components, express it as a magnitude and direction, etc.

Last edited: Mar 24, 2007
3. Mar 24, 2007

### Mentz114

The force between 2 charges is given by

F = K q1 q2/r^2 and the field from a charge is E = K q / r^2

See here
http://en.wikipedia.org/wiki/Electric_field

[edited - thanks Robb ]

Last edited: Mar 24, 2007
4. Mar 24, 2007

### robb_

Careful, you dropped a factor of r in the denominator for the field.