Electric Field between two flat metal plates

[themagiciant95]

Homework Statement

Two large, flat metal plates are held parallel to each other and separated by a distance d.· They are connected together at their edge by a metal strip. A thirt plastic sheet carrying a surface charge \sigma per unit area is placed between the plates at a distance 1/3*d from the upper plate.
Call E1 and E2 the electric field near the upper and lower plates, respectively. What are E1 and E2?

Homework Equations

The Attempt at a Solution

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Im trying to find a good starting point for solving this problem.
First of all,I suppose that the electric field is constant and perpendicular to the plates in every point. But i don't know why. Can you help me ?

 

[haruspex]

the electric field is constant and perpendicular to the plates in every point

For a pair of adjacent plates, yes. Strictly speaking, that is only true for infinite plates, but if you are told the plates are large (in relation to their separation) then that is the expected approximation.
Is that all you need to get started?

 

[themagiciant95]

For a pair of adjacent plates, yes. Strictly speaking, that is only true for infinite plates, but if you are told the plates are large (in relation to their separation) then that is the expected approximation.
Is that all you need to get started?

Ahah, indeed i has not been able yet. I tried to start the problem using the Gauss Law, but i don't know how. And i don't know which are the useful information i can obtain from the problem's text. Can you help me ?

 

[rude man]

You have 2 unknowns, E1 and E2.
You can get one equation from Gauss's law.
You can get the other by relating electric field to potential difference.

 

[Delta2]

Just to help you abit more with @rude man idea,

use as Gaussian surface a rectangle(more accurately a parallelepiped ) that contains the plastic sheet (and its two big opposite sides inside the space that's inbetween the metal plates) and

notice that since the two metal plates are connected with a metal strip they have equal potential $V$. It will be
$E_1\frac{d}{3}=V_p-V=E_2\frac{2d}{3}$ where $V_p$ the potential of the plastic sheet.

 

[themagiciant95]

Just to help you abit more with @rude man idea,

use as Gaussian surface a rectangle(more accurately a parallelepiped ) that contains the plastic sheet (and its two big opposite sides inside the space that's inbetween the metal plates) and

notice that since the two metal plates are connected with a metal strip they have equal potential $V$. It will be
$E_1\frac{d}{3}=V_p-V=E_2\frac{2d}{3}$ where $V_p$ the potential of the plastic sheet.

Is very clear now , thanks.
I have only a doubt, why can we claim that E1 and E2 are constant and the have the same direction ?

 

[Delta2]

Is very clear now , thanks.
I have only a doubt, why can we claim that E1 and E2 are constant and the have the same direction ?

You are right that the fields will not have the same direction. my equation at post #5 needs a minus in front :D.

That they are constant is actually an approximation. If the plates aren't infinite, the fields would depend both on the x (horizontal) and the y(vertical) coordinate (as well the z/depth coordinate). However if the plates are very large when compared to their separation distance then we can treat the plates as infinite in the horizontal dimension (and in the z-depth direction), so the field will not depend on the x-coordinate (neither on z-coordinate), and by applying gauss's law as told you also conclude that the fields will not depend on the y-coordinate either.