Conductive and grounded shells

  • Thread starter Kosta1234
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In summary, the problem involves determining the potential and charge distribution in a system of conductive and grounded shells. The solution involves using Gauss' law to find the electric field in different regions bounded by the shells, with the assumption that the potential is continuous. The area charge density on the edges of the shells can be found using the jump in electric field at those points.
  • #1
Kosta1234
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1
Homework Statement
conductive and grounded shells
Relevant Equations
## E\cdot dS = \frac {q}{\epsilon_0} ##
Problem Statement: conductive and grounded shells
Relevant Equations: ## E\cdot dS = \frac {q}{\epsilon_0} ##

Hi.
I'll be glad if you can help me with this question.I've two conductive and grounded shells with radius 'a' and radius 'b' with their center on the same point.
And another conductive (But not grounded) shell with radius R (a<R<b) and charge density ## \sigma ##. I'm asked to figure out what is the potential in space and the charge distribution on the grounded shells.So my way to solution was first to figure out what is the potential in space is using gaus law.
so where ## r < a ## the Electric field is zero therefore the potential in ## r < a ## is constant and because of the potential on the edge is zero so ## V_{r<a} = 0 ##.

When I'm moving to ## R < r < a ##, (between the grounded conductive shell and the conductive shell) . I'm in a little bit in a trouble here.
grounding the shell will bring the potential to be to zero, and will make not trivial charge distribution on it, but does that mean that in TOTAL all the charge is zero on the conductive grounded shell? So that I could use gaus law here again, and in this way I get that ##E\vec = 0## in ## R < r < a ## as well?

thanks.
 
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  • #2
grounding the shell will bring the potential to be to zero, and will make not trivial charge distribution on it, but does that mean that in TOTAL all the charge is zero on the conductive grounded shell?
No.

Using Gauss' law, you should be able to show that the fields in the two regions (1) ##a < r < R## and (2) ##R < r < b## are inverse square fields. That is, ##E_1 = A/r^2## and ##E_2 = B/r^2##, where A and B are unknown constants. Try to think of a way to set up two equations to solve for ##A## and ##B##.

Hint: The statement that the inner and outer conductors are grounded means that those two conductors are at the same potential. Also, can you use Gauss' law to determine the jump in E as you cross the middle conductor at r = R?
 
  • #3
TSny said:
No.

Using Gauss' law, you should be able to show that the fields in the two regions (1) ##a < r < R## and (2) ##R < r < b## are inverse square fields. That is, ##E_1 = A/r^2## and ##E_2 = B/r^2##, where A and B are unknown constants. Try to think of a way to set up two equations to solve for ##A##

Why are those inverse?
Is thia because the amount of charge that coming to the inner grounded conductive shell?
 
  • #4
Kosta1234 said:
Why are those inverse?
Inverse square, i.e. each field is proportional to the inverse square of the distance, i.e. r-2.
 
  • #5
The flaw in your argument is that you've assumed the Potential, V(r) is continuous (i.e if it is a constant inside the volume of the shell and zero on the surface, then it must be zero inside) which in general need not be the case. ( This problem is a clear counter-example to that).
The way you go about solving this is by assuming some total charge, say q1 and q2 is deposited by the ground on the shells with radii 'a' and 'b' respectively to maintain equilibrium/minimize potential energy. Now you can apply Gauss' Law to find the electric field in the region bounded by the conductors and outside. Now think of a way to solve for q1 and q2.
 
  • #6
Sriram Akella said:
The flaw in your argument is that you've assumed the Potential, V(r) is continuous
Seems like a fair assumption to me. Each shell produces a continuous potential, therefore the sum is continuous.
Sriram Akella said:
find the electric field
There is no need to find any fields. It can all be done with potentials.
 
  • #7
Yeah, you're right. I'm sorry. The Potential is continuous, I stand corrected. And yes, it can all be done with potentials. I'm really sorry about the previous post...
 
  • #8
Thank you I hope I solved it right.

On one of the next questions I've been asked to figure out what is the area charge density on the edges of the grounded shells if the shells were with width ## \Delta a ## and ## \Delta b ##.

How can I figure out this?
I know that the charge density with jump of electric field is:
## \Delta E = \frac {\sigma}{\epsilon _0} ##

But what about the edges?

Edit:
I can figure out easly the electric field in the space, and the Electric field inside the conductor is ## \vec E = 0 ## (even if it's grounded?), and I can know the charge density on the edges using
## \Delta E = \frac {\sigma}{\epsilon _0} ##
 
Last edited:
  • #9
I missed this before...
Kosta1234 said:
does that mean that in TOTAL all the charge is zero on the conductive grounded shell?
No.
Kosta1234 said:
I hope I solved it right.
If you post your solution someone will check it.
 
  • #10
Kosta1234 said:
what is the area charge density on the edges of the grounded shells if the shells were with width Δa and Δb .

How can I figure out this?
Consider a Gaussian surface lying between the inner and outer surfaces of the shell.
 
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1. What is the purpose of a conductive shell?

A conductive shell is designed to protect electronic equipment or devices from electromagnetic interference (EMI) by providing a barrier that redirects or absorbs the electrical energy. It also helps to prevent electrostatic discharge (ESD) from damaging sensitive components.

2. How does a conductive shell work?

A conductive shell works by providing a conductive path for any stray electrical energy to travel through. This path directs the energy away from the equipment or device, preventing it from causing damage. The shell is typically made of a conductive material such as metal, which can absorb or reflect the energy.

3. Is a conductive shell the same as a grounded shell?

No, a conductive shell and a grounded shell are not the same. A conductive shell is designed to redirect or absorb electrical energy, while a grounded shell is connected to an external ground source to dissipate any excess energy. However, a conductive shell can also be grounded for added protection.

4. Do all electronic devices need a conductive shell?

No, not all electronic devices need a conductive shell. Devices that emit a lot of electromagnetic radiation, such as radios or cell phones, may require a conductive shell to reduce interference. However, devices that emit little to no radiation, such as calculators, may not need a conductive shell.

5. Can a conductive shell be used to protect against all types of EMI?

No, a conductive shell may not be effective against all types of EMI. It is designed to protect against radiated EMI, which is caused by electromagnetic waves. However, it may not be as effective against conducted EMI, which is caused by electrical currents traveling through wires or other conductors.

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