Gauss' Law & Surface Charge Density

In summary, the total field inside the metallic plate is found by adding all three fields (including the applied external field). The external field is given by the first equation in your opening post, and the charge density on each sheet is +\sigma,~-\sigma.
  • #1
Shigun
3
0

Homework Statement


A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find:
(a) the charge density of each face of the plate and
(b) the total charge on each face


Homework Equations


[tex]
E = \frac{\sigma}{2\epsilon {0}}
[/tex]

[tex]
\sigma = \frac{q}{a}
[/tex]

The Attempt at a Solution


I missed about a week due to unforeseen circumstances, so I'm struggling with this to catch up, but so far:

Area, converted to meters, is 0.25 m^2. Electric field, E, converted to nano units would be 80x10^-9. My understanding was that [tex]\Phi _E = 2EA[/tex] so [tex]\Phi _E = 2 \times \frac{\sigma}{2\epsilon {0}} \times \frac{q}{a}[/tex] which comes out to [tex]\Phi _E = 2 \times \frac{\sigma}{8.854\times(10^-12)}\times\frac{q}{0.50\times0.50} [/tex], however I'm a bit confused on this. How do I find the value of q, and am I correct in my setup such that [tex]\sigma[/tex] is setup within the second fraction, over 8.854*10^-12?

Any help is greatly appreciated
 
Last edited:
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  • #2
What do you know to be true about a metal (conductor) when it comes to electric fields?
 
  • #3
Gokul43201 said:
What do you know to be true about a metal (conductor) when it comes to electric fields?

Charge is only ever on the surface, and the interior has a 0 charge. I also know, as mentioned in the equation (and from my reading) that a conductors equation is:

[tex]E=\frac{\sigma}{2\epsilon{0}}[/tex]

While a nonconductor is

[tex]E=\frac{\sigma}{\epsilon{0}}[/tex]

I also know that finding the surface density is supposed to involve using a Gaussian cylinder.

Correct?
 
  • #4
Shigun said:
Charge is only ever on the surface, and the interior has a 0 charge.
Correct. Also there is no net electric field in the interior of a conductor. So when a conducting plate is placed in an external electric field, what needs to happen to the positive and negative charges in the conductor for this condition to be satisfied? (Hint: If you place freely moving positive and negative charges in an electric field, what happens to them?)
 
  • #5
Gokul43201 said:
Correct. Also there is no net electric field in the interior of a conductor. So when a conducting plate is placed in an external electric field, what needs to happen to the positive and negative charges in the conductor for this condition to be satisfied? (Hint: If you place freely moving positive and negative charges in an electric field, what happens to them?)

I'm kind of confused by the wording, but my guess of what you are saying is referring to the charges separating out to each side of the plate, one being negative and one being positive. Yes?
 
  • #6
Shigun said:
I'm kind of confused by the wording, but my guess of what you are saying is referring to the charges separating out to each side of the plate, one being negative and one being positive. Yes?
Yes. You can figure out which side will get what kind of charge based on the direction of the external field. But more importantly, each of these two sheets of charge will now produce their own electric fields. The negative sheet will produce field lines towards it and the positive sheet will produce field lines away from it. The field from each of these sheets will be given by the first equation in your opening post.

Inside the metallic plate, the total field is found by adding all three fields (including the applied external field). Can you now write an expression for the total field inside, in terms of the external field (call it Eo) and the charge density on each sheet (say, [itex]+\sigma,~-\sigma[/itex])? Draw a diagram to help yourself account for all the fields with their correct directions.
 

Related to Gauss' Law & Surface Charge Density

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric field at a point to the charge enclosed by a surface surrounding that point. It is represented by the formula ∮SE·dA = Qenc0, where ∮SE·dA is the surface integral of the electric field, Qenc is the enclosed charge, and ε0 is the permittivity of free space.

2. What is surface charge density?

Surface charge density is a measure of the amount of electric charge per unit area on a surface. It is represented by the symbol σ and is defined as σ = Q/A, where Q is the amount of charge and A is the area of the surface. It is often used in Gauss' Law to calculate the electric field at a point near a charged surface.

3. How is surface charge density related to Gauss' Law?

Surface charge density is directly related to Gauss' Law as it is used to calculate the electric field at a point near a charged surface. The surface integral in Gauss' Law (∮SE·dA) is essentially the sum of the electric field contributions from all the surface elements, with each element weighted by its corresponding surface charge density. Therefore, understanding the concept of surface charge density is crucial in applying Gauss' Law to solve problems.

4. What is the unit of surface charge density?

The unit of surface charge density is coulombs per square meter (C/m2). This unit is derived from the definition of surface charge density (σ = Q/A) where charge (Q) is measured in coulombs (C) and area (A) is measured in square meters (m2).

5. Can surface charge density be negative?

Yes, surface charge density can be negative. This means that the surface has a net negative charge and the electric field will point towards the surface. It is important to note that the direction of the electric field is determined by the sign of the surface charge density, not the magnitude. A negative surface charge density does not necessarily mean a weaker electric field compared to a positive surface charge density.

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