# Electrical Field between plate

1. Sep 3, 2011

Problem:

Two square metal plates are placed parallel to each other, separated by a distance d= 1.56 cm. The plates have sides of length L = 0.560 m. One of the plates has charge Q= + 1.74×10-6 C, while the other plate has charge -Q. Calculate the magnitude of the electric field between the plates, not close to the edge, i.e., assume a uniform surface charge distribution

Here's what I have so far: The electric field between them obviously isn't 0.

I've got sigma = (magnitude of charge)/area.
The electric field for one plate is E = sigma/(2 * epsilon).
Since the fields from both plates in between them point in the same direction, the total field would be E = sigma/epsilon. Would I use E=sigma/epsilon OR E=sigma/(2*epsilon) here for the correct answer?

epsilon is just 8.85 here not 8.85E-12 im 99% but I dont really know why.

How in the WORLD do I find the area to find sigma though?

2. Sep 3, 2011

### Staff: Mentor

You want the total field, of course.

You're wrong here: It's 8.85E-12.

The plates are square. Find the area.

3. Sep 5, 2011

Im still not getting it right...

I used 1.74×10-6 C as my charge and found the area of the one plate...do I need to find the total area (ie. add the two areas?)

I then got sigma using the above and plugged it into E= sigma / (2* epsilon)

I tried using 8.85 and 8.85E-12 and both incorrect.

Help?

4. Sep 5, 2011

### Staff: Mentor

Good. What did you get for the area.

No.

What did you get for sigma?

No point in plugging the wrong value.

5. Sep 5, 2011

area came out to: .3136 m^2
sigma came out to be: 5.548E-6

Plugged that into E = sigma / (2* epsilon) gave me: 313472.85 N/C and it was incorrect.

6. Sep 5, 2011

### Staff: Mentor

You want the total field, not just that from one side. Double your answer.

7. Sep 5, 2011

I thought since we were doing 2*epsilon it was for both plates...aka total field.

I have the answer right now but dont really get why we had to double as I thought it was as stated above.

Thanks for the help!

8. Sep 5, 2011

### Staff: Mentor

That factor of 2 is in the denominator, so that version is only half the field. (See my response in post #2.)

9. Sep 5, 2011