What is the Magnitude of the Electric Field Between Parallel Plates?

[badd99]

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 I am 99% but I don't really know why.

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

 

[Doc Al]

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?

You want the total field, of course.

epsilon is just 8.85 here not 8.85E-12 I am 99% but I don't really know why.

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

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

The plates are square. Find the area.

 

[badd99]

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?

 

[Doc Al]

I used 1.74×10-6 C as my charge and found the area of the one plate...

Good. What did you get for the area.

do I need to find the total area (ie. add the two areas?)

No.

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

What did you get for sigma?

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

No point in plugging the wrong value.

 

[badd99]

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.

 

[Doc Al]

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

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

 

[badd99]

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

I have the answer right now but don't really get why we had to double as I thought it was as stated above.

Thanks for the help!

 

[Doc Al]

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

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

 

[badd99]

Thank you very much...I posted one more question if you don't mind.

Much appreciated!