Electric field between two plates

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SUMMARY

The discussion focuses on calculating the electric field and potential between two parallel plates with uniform charge densities of +ρ and -ρ. The electric field (E) is derived using Gauss's Law, leading to the conclusion that E = ρ*2w/ε₀, where ε₀ is the vacuum permittivity. The participant confirms the correctness of their expression for the charge element (ΔQ) and recognizes that integrating the electric field from -infinity to +infinity results in E = 0, which is consistent with the symmetry of the problem. The discussion emphasizes the efficiency of Gauss's Law over the Biot-Savart Law for symmetric configurations.

PREREQUISITES
  • Understanding of Gauss's Law and its application in electrostatics
  • Familiarity with electric field and electric potential concepts
  • Knowledge of charge density and its implications in electric fields
  • Basic calculus for integration in physics problems
NEXT STEPS
  • Study the application of Gauss's Law in various symmetrical charge distributions
  • Learn about electric potential calculations in electrostatics
  • Explore the differences between Gauss's Law and the Biot-Savart Law
  • Investigate the implications of vacuum permittivity in electric field calculations
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Physics students, electrical engineers, and anyone interested in understanding electrostatics and electric fields in symmetrical charge configurations.

Robin Lee
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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.
 
Last edited:
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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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