What is the magnitude of the electric field between the two plates?

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Homework Help Overview

The problem involves calculating the magnitude of the electric field between two charged plates, specifically two square plates with given dimensions and charge values. The context is within electrostatics, focusing on electric fields generated by charged objects.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of charge density and its application in the electric field formula. Questions arise regarding the correct interpretation of the charge on the plates and how to approach the problem using Gauss's Law. There is also exploration of the effects of oppositely charged versus similarly charged plates on the electric field.

Discussion Status

The discussion is active, with participants offering insights into the relationship between charge density and electric field calculations. Some guidance has been provided regarding the assumptions of uniform charge distribution and the implications of plate charge types on the electric field. There is no explicit consensus yet, as participants are still clarifying their understanding and calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the depth of exploration. There is also a mention of potential confusion regarding the application of formulas and the interpretation of charge values.

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Homework Statement


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?



Homework Equations


E={sigma}{epsilon_0} .


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?
 
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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.))
 
Last edited:
so would it just be (10x10)/10x10^-9 to get the charge density then divide that by E_0 to get the answer?
 
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.
 
oh opps...it was suppose to 12x10^-9 which is the charge given in the problem
 

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