Difference between These Electric Field Formulas

In summary, there are two formulas for electric fields produced by uniformly charged parallel plates. One formula is for an infinite plane, while the other is for a plate with a definite length, such as in a parallel plate capacitor. The difference between the two formulas is that one is for a single sheet, while the other is for the field between two oppositely charged sheets. It is important to note that in both cases, the sheets are considered to be very large compared to the distance away from the sheet.
  • #1
wahaj
156
2
just a quick question. I seem to have come across two very similar formulas for electric fields produced by uniformly charged parallel plates. My book gives the formula as
[tex] E = \frac{\omega}{\epsilon_0} [/tex]
on the other hand the notes from my prof tell me that
[tex] E = \frac{\omega}{2\epsilon_0} [/tex]

My book or the notes don't mention the other formulas and this whole concept is still new to me to I'm a bit confused. what is the difference between the two formulas?
 
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  • #3
so basically the first equation is for an infinite plane while the second one is for a plate with a definite length like in a parallel plate capacitor. Did I get that right?
 
  • #4
Not quite. One formula (with the "2" in the denominator) is for a single sheet, the other is for the field between two oppositely charged sheets. One way to remember this is that having two sheets produces twice the field, which gets rid of the "2" in the formula.

In both cases the sheets are considered to be very large compared to the distance away from the sheet.
 
  • #5


The difference between the two formulas lies in the value of the constant used for the permittivity of free space, denoted by ε0. In your book, the formula uses the value of ε0 as it is, while in your professor's notes, the formula uses the value of 2ε0.

This difference in the value of ε0 may seem minor, but it has a significant impact on the resulting electric field strength. This is because ε0 is a fundamental constant that represents the ability of the medium to permit the flow of electric charge. A higher value of ε0 means that the medium has a higher capacity to permit the flow of charge, resulting in a stronger electric field.

Therefore, the formula in your book may be more accurate as it uses the exact value of ε0, while the formula in your professor's notes may be an approximation that takes into account other factors. It is best to clarify with your professor which formula to use in your specific case to ensure accuracy in your calculations.

Additionally, it is important to note that there may be other factors that can affect the electric field produced by uniformly charged parallel plates, such as the distance between the plates and the magnitude of the charge on the plates. These factors may also be addressed in your professor's notes, so it is important to refer to them for a more comprehensive understanding of the concept.
 

FAQ: Difference between These Electric Field Formulas

What is the difference between Coulomb's Law and Gauss's Law?

Coulomb's Law and Gauss's Law are both used to calculate the electric field, but they differ in their approach. Coulomb's Law relates the electric field to the distance between two point charges, while Gauss's Law calculates the electric field by considering the total charge enclosed by a surface.

Can Gauss's Law be used for all types of charge distributions?

Gauss's Law can be used for any type of charge distribution, as long as the charge is enclosed by a closed surface. However, for complex charge distributions, it may be easier to use Coulomb's Law to calculate the electric field.

How do the electric field formulas differ for point charges and continuous charge distributions?

The formula for the electric field for a point charge is given by Coulomb's Law, while for a continuous charge distribution, it is given by Gauss's Law. Additionally, the electric field for a point charge is a vector quantity, while for a continuous charge distribution, it is a scalar quantity.

Are there any limitations to using Coulomb's Law or Gauss's Law to calculate the electric field?

Coulomb's Law and Gauss's Law are both valid for static electric fields. They may not accurately predict the electric field in situations involving moving charges or changing electric fields, such as in electromagnetic waves.

Can the electric field be negative?

Yes, the electric field can be negative. This indicates that the direction of the electric field is opposite to the direction of the force that would be exerted on a positive test charge placed in that field.

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