Using Gauss's Law on two separated, and infinite plates

In summary: So the electricfield is zero between the two parallel sheets because the field lines would have to go throughboth sheets and cancel out.Review a Gaussian pillbox for a (single) non-conducting sheet and the derivation of the electricfield on either side of the sheet.Now, as you have surmised, there can be no field between the two parallel sheets.Now consider the fact that electric field lines originate on positive charges and endon negative charges (they have to go somewhere). So the electricfield is zero between the two parallel sheets because the field lines would have to go throughboth sheets and cancel out.
  • #1
erick rocha
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1

Homework Statement


The figure shows cross-sections through two large, parallel, nonconducting sheets with identical distributions of positive charge with surface charge density σ = 1.06 × 10-22 C/m2. What is the y component of the electric field at points (a) above the sheets, (b) between them, and (c) below them?

Homework Equations



So since they have equal surface density, and their postive polarities are opposite of each other, I would think that they electric fields would just cancell each other out right? Appereantly not. Their elecctric field hace equal magnitude in opposite direction. Why would the doward compnent cancel the upward component?

The Attempt at a Solution


The E field in the in between the plates is zero since the inside component cancel each other out.
On the ouside...
E(net)= E(top)-E(bottom)
=σ/2ε-σ/2ε
=0
But the answer is actually
=σ/ε
 

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  • #2
Hello erick, :welcome:

I wonder why you write E(net)= E(top)-E(bottom) instead of E(net)= E(top) + E(bottom). Are you aware the ##\vec E## field has a direction ? I think you are: after all, you let them cancel in the space between the plates. So what are the directions above both plates ? And below both plates ?
 
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  • #3
BvU said:
Hello erick, :welcome:

I wonder why you write E(net)= E(top)-E(bottom) instead of E(net)= E(top) + E(bottom). Are you aware the ##\vec E## field has a direction ? I think you are: after all, you let them cancel in the space between the plates. So what are the directions above both plates ? And below both plates ?
Thanks for the welcome and the response BuV!
Conventionally we say that the E Field goes from positive to negative and since there is no field inside the object that must mean the E field vector from the top plate with a positively charge surface a must be pointing towards the positive y-axis towards infinity. Also the bottom plate of the bottom surface, that is also positively charged, must be pointing towards the negative y-axis towards infinity. The same logic as why the E field vectors in the middle canceled out.
 
  • #4
erick rocha said:
Conventionally we say that the E Field goes from positive to negative and since there is no field inside the object that must mean the E field vector from the top plate with a positively charge surface a must be pointing towards the positive y-axis towards infinity
How about reasoning the other way around: E-field points away from positive charge, therefore the two (identical magnitude) contributions cancel in between and add up outside both plates !
 
  • #5
Review a Gaussian pillbox for a (single) non-conducting sheet and the derivation of the electric
field on either side of the sheet.
Now, as you have surmised, there can be no field between the two parallel sheets.
Now consider the fact that electric field lines originate on positive charges and end
on negative charges (they have to go somewhere).
 

FAQ: Using Gauss's Law on two separated, and infinite plates

1. What is Gauss's Law and how does it apply to two separated, infinite plates?

Gauss's Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. In the case of two separated, infinite plates, Gauss's Law can be used to calculate the electric field between the plates and determine the charge distribution on the plates.

2. How do you calculate the electric field between two separated, infinite plates using Gauss's Law?

To calculate the electric field between two separated, infinite plates, you can use Gauss's Law by drawing a Gaussian surface between the plates and using the enclosed charge and permittivity of free space to solve for the electric field. This can be simplified for infinite plates by assuming the electric field is constant between the plates.

3. Can Gauss's Law be applied to any type of charge distribution on the plates?

Yes, Gauss's Law can be applied to any type of charge distribution on the plates as long as the charge is enclosed by the Gaussian surface. This means that the charge must be contained within the surface and not extend beyond it.

4. What is the significance of using infinite plates in Gauss's Law calculations?

Using infinite plates allows for simplification in the calculations, as it assumes that the electric field is constant between the plates. This is a reasonable assumption for very large plates, and it allows for easier calculations and a more accurate result.

5. Are there any limitations to using Gauss's Law on two separated, infinite plates?

One limitation is that Gauss's Law assumes a uniform charge distribution on the plates. If the charge distribution is not uniform, the results may be less accurate. Additionally, the plates must be truly infinite in size for the calculations to be accurate, which may not always be the case in real-world scenarios.

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