Electric Field of a Conducting Slab

In summary, the problem involves finding the electric field, E(r), at a general point r above a horizontal conducting slab with a charge Q placed at height b. The electric field above a plane is given by sigma*r/2*epsilon_zero*r, but the thickness of the slab may affect this calculation. The appropriate boundary conditions for this problem include continuity at the surface of the slab and zero at infinity. One can assume the thickness of the slab is negligible and consider it an equipotential plane. The electric field induced on the slab will have the same magnitude as the point charge above it, but the geometry of the slab may affect the field. The method of image charges may be used to solve this problem.
  • #1
aurora14421
24
0

Homework Statement



A charge Q is placed at height b abouve a plane horizontal conducting slab.

Write down the electric field, E(r), at a general point r above the slab (taking the point r=0 to be the point on the slab directly beneath the charge), and show that it satisfies the appropriate boundary conditions.

Homework Equations





The Attempt at a Solution



Ok, I know that the electric field above a plane is sigma*r/2*epsilon_zero*r, but since this slab has a thickness where there will be no electric field, does this make a difference? Can I still use Gauss' Theorem? And what does it mean by "appropriate boundary conditions"? Just that it's continuous at the surface of the slab and zero at infinity?

Just, generally, where do I start with this question?

Thanks
 
Physics news on Phys.org
  • #2
For this type of problem one usually assumes that the thickness of the slap is negligible. In other words, one can consider the slab an equipotential plane.

Note that since this slab is a conductor an electric field will be induced on this plane. What can you say about this electric field?
 
  • #3
Is this supposed to be solved by the method of image charges, or that is something completely different?

Well the charge induced at the surface of the slab will have the same magnitude as the point charge above it. But i am not sure what that tells me about the field since the geometry of the slab is different then the one of the point charge obviously

( I have problems with something very similar so thought to jump in if that's ok.. )
 

Related to Electric Field of a Conducting Slab

1. What is the electric field of a conducting slab?

The electric field of a conducting slab refers to the distribution of electric charges on the surface of a flat, metallic slab. This field is created by the presence of electric charges, and it is responsible for the movement of these charges within the slab.

2. How is the electric field of a conducting slab calculated?

The electric field of a conducting slab can be calculated using the Gauss's Law, which states that the electric flux through any closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space. Alternatively, it can also be calculated by using the formula E = σ/ε, where σ is the surface charge density and ε is the permittivity of the material.

3. What factors affect the electric field of a conducting slab?

The electric field of a conducting slab is affected by several factors, including the thickness of the slab, the surface charge density, and the permittivity of the material. Additionally, the presence of other nearby electric charges or conductors can also influence the electric field.

4. Can the electric field of a conducting slab be changed?

Yes, the electric field of a conducting slab can be changed by altering the factors that affect it, such as the surface charge density or the permittivity of the material. Additionally, the application of an external electric field can also alter the distribution of charges on the surface of the slab.

5. What is the significance of the electric field of a conducting slab?

The electric field of a conducting slab plays a crucial role in understanding the behavior of electric charges and the movement of current in conducting materials. It is also essential in various applications, such as in the design and functioning of electronic devices and circuits.

Similar threads

Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Electromagnetism
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
843
  • Advanced Physics Homework Help
Replies
4
Views
3K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Replies
16
Views
8K
  • Advanced Physics Homework Help
Replies
26
Views
4K
Back
Top