Electric field between two plates

AI Thread Summary
The discussion focuses on calculating the electric field and potential between two parallel plates with uniform charge distribution. The user correctly identifies the charge element Delta-Q and applies Gauss's Law using a "pillbox" approach to derive the electric field. They express the electric field as E = ρ*2w/ε₀, confirming the units are consistent. The user acknowledges that Gauss's Law simplifies the problem compared to other methods. The conversation highlights the effectiveness of Gauss's Law in solving symmetric charge distribution problems.
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1. Positive and negative charges are distributed evenly in two parallel plates with width "w". The density of the positive and negative charge is +\rho and -\rho. Assume that the length of the plates at y direction is \infty

(a) Find the electric field in the plates
(b) Let the electric potential at x=-w equal to 0, find the electric potential in the plates.



Homework Equations


\DeltaE=k*Delta-Q*r/r2 where r is the unit vector -- (1)


3. My approach
I sliced an infinitesimal portion of the plate horizontally denoted as Delta-y and represent the charge Delta-Q = rho *Delta-y*2*w. Is my expression for Delta-Q correct?

r is the observation point which is at opposite to the portion Delta-y on the negative plate of which magnitude is "w".

Substitute my expression for Delta-Q into (1), I obtain Delta-E = k*rho/w.

Do I have to integrate Delta-E from -infinity to +infinity to obtain the electric field in the plates? If so, I will get E=0. Does this make sense?

Thank you.
 
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This is a Gauss's Law problem. Use the standard "pillbox" with its faces perpendicular to the x-axis.
 
kuruman said:
This is a Gauss's Law problem. Use the standard "pillbox" with its faces perpendicular to the x-axis.

I obtained \rho*2w/vacuum permittivity. I also did a sanity check that the unit is correct.

Thanks for the guidance. Now I learned that for highly symmetric problem, Gauss's Law really makes the problem much easier than using Biot-Savart Law.
 
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