Understanding Electric Field of Positive & Negative Plates

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SUMMARY

The electric field generated by a single charged plate is defined as ##E = \frac{\sigma}{2\epsilon_0}##, where ##\sigma## represents the surface charge density and ##\epsilon_0## is the permittivity of free space. When considering both a positive and a negative plate, the total electric field is not simply doubled; instead, it remains ##\vec E = \frac{\sigma}{\epsilon_0}## due to the superposition of the fields from both plates. This conclusion is derived using a pillbox Gaussian surface to analyze the electric field behavior in the vicinity of the plates.

PREREQUISITES
  • Understanding of electric fields and charge distribution
  • Familiarity with Gaussian surfaces and their applications in electrostatics
  • Knowledge of the permittivity of free space (##\epsilon_0##)
  • Basic concepts of superposition principle in physics
NEXT STEPS
  • Study the application of Gaussian surfaces in electrostatics
  • Explore the concept of electric field lines and their representation
  • Learn about the effects of multiple charged plates on electric fields
  • Investigate the relationship between surface charge density and electric field strength
USEFUL FOR

Students of physics, educators teaching electrostatics, and anyone seeking to deepen their understanding of electric fields generated by charged plates.

annamal
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Homework Statement
Two large conducting plates carry equal and opposite charges, with a surface charge density as shown in image. The separation between the plates is d. What is the
electric field between the plates?
Relevant Equations
##\vec E = \frac{\sigma}{\epsilon_0}##
I am confused with the solution. It says ##\vec E = \frac{\sigma}{\epsilon_0}##. Shouldn't E = ##2*\vec E = 2*\frac{\sigma}{\epsilon_0}##? Electric field of the positive plate and electric field of the negative plate.

Screen Shot 2022-04-25 at 11.02.53 PM.png
 
Last edited:
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annamal said:
I am confused with the solution. It says ##\vec E = \frac{\sigma}{\epsilon_0}##. Shouldn't E = ##2*\vec E = 2*\frac{\sigma}{\epsilon_0}##?
It should not. Use a pillbox Gaussian surface with one flat face inside the conductor and the other flat face in the vacuum to see why and become unconfused.
 
Well if you mean that the total E-field should be double of that of the one plate, yes that is true but the field of one plate is ##E=\frac{\sigma}{2\epsilon_0}##. So the total field is $$2\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$$
 
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