What is the electric field above an infinite plate with a given charge?

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SUMMARY

The electric field above an infinite charged plate can be calculated using the formula E = σ / (2ε₀) for a single charged plate, where σ is the charge density and ε₀ is the permittivity of free space. In this discussion, a metal plate measuring 8 cm on each side carries a total charge of 60 microC, leading to a calculated electric field of 53 MN/C at a height of 0.5 mm above the plate's surface. The confusion arose from the incorrect application of the formula for a conductor instead of recognizing the single plate scenario, which requires halving the charge density in the calculation.

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  • Understanding of electric fields and charge density
  • Familiarity with the concepts of conductors and insulators
  • Knowledge of the permittivity of free space (ε₀)
  • Basic principles of electrostatics and capacitors
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  • Study the derivation and application of the electric field equations for charged plates
  • Learn about the differences between conductors and insulators in electrostatics
  • Explore the concept of parallel plate capacitors and their electric fields
  • Investigate the role of charge density in calculating electric fields
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manenbu
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Homework Statement



From Resnick and Halliday:

A metal plate 8 cm on a side carries a total charge of 60 microC. Using the infinite plate approximation, calculate the electric field 0.5 mm above the surface of the plate near the plate's center.

Homework Equations



(1) E = \frac{\sigma}{\epsilon_0}
or
(2) E = \frac{\sigma}{2\epsilon_0}
\sigma being charge density.

The Attempt at a Solution



I got a bit confused here. Equation (1) should be used when the plate is a conductor, and equation (2) should be used when the plate is an insulator, according to the explanation in the book.
However, when I use equation (1) to calculate, I get 106MN/C, and in the answers in the back of the book it says it should be 53 MN/C. Either I used the wrong equation, or I should have used half of the charge.
What went wrong?

And honestly, I really did not understand why there are 2 different equations for each situation (conductor, insulator). I tried deriving them on my own but no success in gaining real understanding, so I'd be really glad if someone could shed some light on the subject.
 
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manenbu said:
A metal plate 8 cm on a side carries a total charge of 60 microC. Using the infinite plate approximation, calculate the electric field 0.5 mm above the surface of the plate near the plate's center.
I calculate an answer that differs by a factor of 10 from the given answer. Are you sure that the problem statement specifies 60 μC instead of say, 6.0 μC?

Homework Equations



(1) E = \frac{\sigma}{\epsilon_0}
or
(2) E = \frac{\sigma}{2\epsilon_0}
\sigma being charge density.

The Attempt at a Solution



I got a bit confused here. Equation (1) should be used when the plate is a conductor, and equation (2) should be used when the plate is an insulator, according to the explanation in the book.
However, when I use equation (1) to calculate, I get 106MN/C, and in the answers in the back of the book it says it should be 53 MN/C. Either I used the wrong equation, or I should have used half of the charge.
What went wrong?

And honestly, I really did not understand why there are 2 different equations for each situation (conductor, insulator). I tried deriving them on my own but no success in gaining real understanding, so I'd be really glad if someone could shed some light on the subject.
I'm not familiar with Resnick and Halliday text. However, I doubt a textbook would make such a mistake.

Equation (1) applies to the situation where you have two separate, oppositely charged plates, and you're looking for the electric field in between them. This is the situation when working with a parallel plate capacitor (without a dielectric).

Equation (2) applies to a single charged plate.

The difference in application between the equations has nothing really to do with the plate(s) being conducting or insulating. Rather its a matter of whether you have one or two plates.

Capacitors will invariably be constructed of conducting plates (otherwise it wouldn't function as a capacitor). Maybe that's from where the confusion stems.
 

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