Electric field inside a flat sheet of copper

In summary, the electric field at the surface of a conductor is always given by E=\frac{\sigma}{\epsilon_{0}}\hat{n}, and for a sheet with surface charge density on one face, the electric field just outside of each face takes the form E=\frac{\sigma}{2\epsilon_{0}}\hat{n}. These fields cancel each other inside the conductor but add up outside, resulting in a total electric field of E=\frac{\sigma}{\epsilon_{0}}\hat{n}. For a thin sheet of copper, the electric field inside may not be zero due to the maximum number of electrons per unit volume. However, for practical fields and thicknesses, this is not a concern.
  • #1
Demon117
165
1
Suppose that you have a sheet of copper that carries a surface charge of [itex]\sigma[/itex] on each face. The electric field just outside of each face can be calculated by plane symmetry and the solution would take the form

[itex]E=\frac{\sigma}{2\epsilon_{0}}\hat{n}[/itex]

correct?

If not I have to go back and rework this. Also, wouldn't the electric field inside the sheet be zero? The reason would be that the electric flux through the surface is zero.
 
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  • #2
Correct and yes. E is always zero inside a conductor.
 
  • #3
The electric field at the surface of a conductor is always:

[itex]E=\frac{\sigma}{\epsilon_{0}}\hat{n}[/itex]

Each charged face produces an electric field given by:

[itex]E=\frac{\sigma}{2\epsilon_{0}}\hat{n}[/itex]

These fields cancel each other inside the conductor (they are in opposite directions), but they are in the same direction outside: we have to add them:

[itex]E=\frac{\sigma}{2\epsilon_{0}}\hat{n}+\frac{\sigma}{2\epsilon_{0}}\hat{n} = \frac{\sigma}{\epsilon_{0}}\hat{n}[/itex]


matumich26 said:
Suppose that you have a sheet of copper that carries a surface charge of [itex]\sigma[/itex] on each face. The electric field just outside of each face can be calculated by plane symmetry and the solution would take the form

[itex]E=\frac{\sigma}{2\epsilon_{0}}\hat{n}[/itex]

correct?

If not I have to go back and rework this. Also, wouldn't the electric field inside the sheet be zero? The reason would be that the electric flux through the surface is zero.
 
  • #4
Now it's correct. One should also say that the electric field inside a conductor is only necessarily 0 if you consider only static fields and charge distributions. For time-dependent fields and charge-current distributions, the electric field is not necessarily 0 inside conductors.
 
  • #5
YPelletier said:
The electric field at the surface of a conductor is always:

[itex]E=\frac{\sigma}{\epsilon_{0}}\hat{n}[/itex]

Each charged face produces an electric field given by:

[itex]E=\frac{\sigma}{2\epsilon_{0}}\hat{n}[/itex]

These fields cancel each other inside the conductor (they are in opposite directions), but they are in the same direction outside: we have to add them:

[itex]E=\frac{\sigma}{2\epsilon_{0}}\hat{n}+\frac{\sigma}{2\epsilon_{0}}\hat{n} = \frac{\sigma}{\epsilon_{0}}\hat{n}[/itex]

So the total electric field outside the surface is [itex]E=\frac{\sigma}{\epsilon_{0}}\hat{n}[/itex]? But just outside each surface it is [itex]E=\frac{\sigma}{2\epsilon_{0}}\hat{n}[/itex]?

If that is the case then wouldn't the Electric field be simply [itex]E=\frac{\sigma}{2\epsilon_{0}}\hat{n}[/itex] for the sheet with surface charge density on one face only?
 
  • #6
You have to solve the boundary-value problem properly. It's clear that the field inside the conductor must be 0. Outside of the conductor it's a gradient field. Let's assume the sheet is parallel to the xy-plane. The boundaries may be at [itex]z=\pm d/2[/itex] Due to symmetry the potential can only depend on z. Then you have

[tex]\Delta \Phi(z)=\Phi''(z)=0 \; \Rightarrow \; \Phi(z)=A z[/tex]

with an integration constant [itex]A[/itex]. I've set the other integration constant which just adds to [itex]\Phi[/itex] arbitrarily to 0 since it has no physical significance anyway. Since the field vanishes inside the conductor, you have more precisely

[tex]\Phi(z)=\begin{cases} 0 & \text{for} \quad -d/2<z<d/2 \\
A_> z & \text{for} \quad z \geq d/2 \\
A_< z & \text{for} \quad z \leq -d/2
\end{cases}[/tex]

To determine the A 's you need to know that the jump of the normal component of the electric field, [itex]E_z[/itex] obeys the condition at [itex]z=d/2[/itex]

[tex]\vec{n} \cdot [\vec{E}(d/2+0^+)-\vec{E}(d/2-0^+)]=\frac{\sigma}{\epsilon_0},[/tex]

where [itex]\vec{n}[/itex] is the normal vector pointing outside, i.e., here [itex]\vec{n}=\vec{e}_z[/itex]. From this jump condition you find from [itex]\vec{E}=-\vec{\nabla} \Phi[/itex]

[tex]A_>=-\frac{\sigma}{\epsilon_0}.[/tex]

In the same way you find at the surface at [itex]z=-d/2[/itex] (note that here [itex]\vec{n}=-\vec{e}_z[/itex])

[tex]A_<=+\frac{\sigma}{\epsilon_0}.[/tex]

The solution thus reads

[tex]\Phi(z)=\begin{cases}
0 & \text{for} \quad -d/2<z<d/2, \\
-\frac{\sigma}{\epsilon_0} |z| & \text{for} \quad |z| \geq d/2.
\end{cases}[/tex]

The electric field is

[tex]\vec{E}(z)=-\vec{\nabla} \Phi(z)=\begin{cases}
0 & \text{for} \quad |z|<d/2, \\
\mathrm{sign}(z) \frac{\sigma}{\epsilon_0} \vec{e}_z & \text{for} \quad |z| \geq d/2.
\end{cases}[/tex]
 
  • #7
If you have a really thin sheet of copper and really large fields the electric field inside need not be zero.

There is a maximum number of electrons per unit volume. For practical fields and thicknesses this normally does not matter, but for a really thin sheet of copper there may not be enough free electrons to completely block a field.

I did a calculation and a monoatomic layer of copper could support fields of the order of 1e9 V/m - so forget my argument.
 
  • #8
clem said:
E is always zero inside a conductor.
Provided there is no moving charge of course
 
  • #9
mitch_1211 said:
Provided there is no moving charge of course
I took the original question to be about electrostatics.
 
  • #10
To clarify, this is an electrostatics problem. Thank you all for your support.
 

1. What is an electric field?

An electric field is a physical quantity that describes the influence of electric forces on a charged object. It is represented by vectors that indicate the direction and magnitude of the force on a positive test charge at any given point in space.

2. How is an electric field created inside a flat sheet of copper?

An electric field inside a flat sheet of copper is created by the flow of electrons through the material. When a voltage is applied across the sheet, the electrons move from the negative terminal to the positive terminal, creating an electric field in the direction of the current flow.

3. What factors affect the strength of the electric field inside a flat sheet of copper?

The strength of the electric field inside a flat sheet of copper is affected by the voltage applied across the sheet, the distance between the two terminals, and the thickness of the copper sheet. A higher voltage, shorter distance, and thinner sheet will result in a stronger electric field.

4. How does the electric field inside a flat sheet of copper vary with distance from the surface?

The electric field inside a flat sheet of copper decreases with distance from the surface. This is because the electric field lines spread out as they move away from the sheet, resulting in a weaker field further from the surface.

5. Can the electric field inside a flat sheet of copper be shielded?

Yes, the electric field inside a flat sheet of copper can be shielded by placing a conducting material, such as another sheet of copper, between the original sheet and the source of the electric field. This is due to the fact that the conducting material will redistribute the electric charges and cancel out the electric field inside the shielded area.

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