Electric fields and conductors

In summary, the electric field created by a conductor at a point extremely close to it can be broken down into two components, one created by a tiny element of the conductor approximated as an infinite plane and the other by the rest of the conductor. When applying Gauss' law with a cylinder as the Gaussian surface, the part of the cylinder inside the conductor will not see any electric field, resulting in a discrepancy in the calculation of the electric field. However, for a cylinder, the second component of the electric field can be solved for, resulting in a corrected value of ##\frac{\sigma}{2\epsilon_0}##.
  • #1
archaic
688
214
Hello guys!
The electric field created by a conductor at a point $M$ extremely close to it is ##\vec{E}=\vec{E_1}+\vec{E_2}## where ##\vec{E_1}## is the electric field created by such a tiny bit of the conductor that we can suppose it to be a plane, and since ##M## is extremely close to the conductor such that the distance is really small compared to the size of the plane we further ahead assimilate it to an infinite plane and hence ##\vec{E_1}=\frac{\sigma}{2\epsilon_0}## and this is where I block, when we use Gauss' law on an infinite plane we also account for the electric fields on the other side of the cylinder (here our gaussian surface) i.e the part inside the conductor, but in the case of the conductor the electric field inside of it would be ##\vec{0}## and so ##\vec{E_1}## should be ##\frac{\sigma}{\epsilon_0}## (##E \pi r^2 + 0 + 0= \frac{\sigma \pi r^2}{\epsilon_0}##)
(I have no idea why latex isn't processing this ^)

I cannot see where I've gone wrong.
 
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  • #2
I think you are trying to put a Gaussian surface around a conductor that has a surface charge density ## \sigma ##. Applying Gauss' law, you get ## \\## ## E (2 \pi r)L=\frac{\sigma (2 \pi r L)}{\epsilon_o} ## so that ## E=\frac{\sigma}{\epsilon_o} ##, (pointing radially outward). ## \\ ## This should come as no surprise, even though for an infinite sheet of charge ## E=\frac{\sigma}{2 \epsilon_o} ##.
 
  • #3
Charles Link said:
I think you are trying to put a Gaussian surface around a conductor that has a surface charge density ## \sigma ##. Applying Gauss' law, you get ## \\## ## E (2 \pi r)L=\frac{\sigma (2 \pi r L)}{\epsilon_o} ## so that ## E=\frac{\sigma}{\epsilon_o} ##, (pointing radially outward). ## \\ ## This should come as no surprise, even though for an infinite sheet of charge ## E=\frac{\sigma}{2 \epsilon_o} ##.
Not exactly that, Coulomb's theorem states that the electric field created by a conductor on a point M extremely close can represented as the sum of the field created by an element of surface (which the point is so close to and thus can be "seen" as an infinite sheet of charge density ##\sigma## resulting in ##
\vec{E_1}=\frac{\sigma}{2\epsilon_0}## ) and the rest of the conductor. My problem is with the field created by that one tiny element we've assimilated to be an infinite sheet because when applying Gauss' law with a cylinder as the Gaussian surface, the part of the cylinder that is *inside* the conductor will see no electric field and so, as I see it, ##E_1 \pi r^2 + 0 + 0 = \frac{\sigma \pi r^2}{\epsilon_0}## whereas this should be ##\vec{E_1}=\frac{\sigma}{2\epsilon_0}## apparently
 
  • #4
archaic said:
Not exactly that, Coulomb's theorem states that the electric field created by a conductor on a point M extremely close can represented as the sum of the field created by an element of surface (which the point is so close to and thus can be "seen" as an infinite sheet of charge density ##\sigma## resulting in ##
\vec{E_1}=\frac{\sigma}{2\epsilon_0}## ) and the rest of the conductor. My problem is with the field created by that one tiny element we've assimilated to be an infinite sheet because when applying Gauss' law with a cylinder as the Gaussian surface, the part of the cylinder that is *inside* the conductor will see no electric field and so, as I see it, ##E_1 \pi r^2 + 0 + 0 = \frac{\sigma \pi r^2}{\epsilon_0}## whereas this should be ##\vec{E_1}=\frac{\sigma}{2\epsilon_0}## apparently
##E_2 ## is more difficult to solve for, but in the case of a cylinder, ir must be ##E_2=\frac{\sigma}{2 \epsilon_o} ##.
 

Related to Electric fields and conductors

1. What is an electric field?

An electric field is a region in which an electric charge experiences a force. Electric fields are created by electric charges and can be either positive or negative.

2. How do conductors interact with electric fields?

Conductors are materials that allow electric charges to flow freely. In the presence of an electric field, the charges within a conductor will redistribute themselves until the electric field inside the conductor is zero.

3. What is the difference between an insulator and a conductor?

An insulator is a material that does not allow electric charges to flow freely, while a conductor is a material that does allow electric charges to flow freely.

4. How is the strength of an electric field measured?

The strength of an electric field is measured by its electric field intensity, which is defined as the force per unit charge at a given point in the electric field.

5. Can electric fields pass through conductors?

Yes, electric fields can pass through conductors. However, the presence of a conductor can significantly alter the distribution of the electric field within the surrounding space.

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