electric field between two plates


by tag16
Tags: electric, field, plates
tag16
tag16 is offline
#1
Sep24-09, 06:49 PM
P: 98
1. The problem statement, all variables and given/known data
Two 10 cm x 10cm square plates that are charged with 12 nC and are 6 mm apart. What is the magnitude of the electric field between the two plates?



2. Relevant equations
E={sigma}{epsilon_0} .


3. The attempt at a solution

I know I have to use this equation but I'm not sure how to find the charge density. Also would it be E=sigma/2E_0 , since it's 2 plates or do I just plug the numbers into the formula?
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
TwoTruths
TwoTruths is offline
#2
Sep24-09, 06:59 PM
P: 37
Assuming the plates are charged uniformly, the charge density is simply a direct application of its definition (charge per unit area). So, sigma = total charge / surface area of plate.

Note also that the distance between the two plates is very small compared to the size of the plates. I believe you can treat them as "infinitely" long plates, which will make the below physics easier.

Are you saying the plates are oppositely charged? Or are both - or both +?

Either way, consider the electric field lines of one plate. I will leave the general formula derivation for the electric field to you (hint: Gauss's Law). You should come up with a nice, simple formula.

Now, if we add a plate of opposite charge, what will happen to the field lines? You can see that the field lines outside of the plates disappears! Inside, the magnitude of the electric field will double. Nice, right?

If you add a plate of same charge, what will happen now? Basically the opposite, right? Inside, the two repulsion fields should cancel. Outside, the two repulsion forces should add up to double the original magnitude! Also nice!

Hope this helps!

((Easy way out: Calculate sigma and plug into your equation in section 2. I recommend doing the derivation as it's easy and will help you understand what's going on.))
tag16
tag16 is offline
#3
Sep24-09, 07:20 PM
P: 98
so would it just be (10x10)/10x10^-9 to get the charge density then divide that by E_0 to get the answer?

TwoTruths
TwoTruths is offline
#4
Sep24-09, 07:29 PM
P: 37

electric field between two plates


Depends on the charge of the two plates. I'm assuming they're oppositely charged.

What you have ( [tex]\frac{10*10}{10*10^-9}[/tex] ) is not correct. I don't know where the denominator came from (enlighten?), but the numerator should be on the bottom. It's [tex]\frac{total charge}{total area}=\sigma[/tex].

Sorry, I just figured out this LaTeX thing, and it's pretty cool.
tag16
tag16 is offline
#5
Sep24-09, 07:36 PM
P: 98
oh opps...it was suppose to 12x10^-9 which is the charge given in the problem


Register to reply

Related Discussions
Electric Field: non parallel plates? Introductory Physics Homework 5
Parallel Plates and Electric Field Introductory Physics Homework 1
electric field due to || CONDUCTING plates Introductory Physics Homework 2
Electric Field Between Plates Introductory Physics Homework 9
The electric field of two plates Introductory Physics Homework 9