Electric field inside a flat sheet of copper

[Demon117]

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

$E=\frac{\sigma}{2\epsilon_{0}}\hat{n}$

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.

 

[Meir Achuz]

Correct and yes. E is always zero inside a conductor.

 

[YPelletier]

The electric field at the surface of a conductor is always:

$E=\frac{\sigma}{\epsilon_{0}}\hat{n}$

Each charged face produces an electric field given by:

$E=\frac{\sigma}{2\epsilon_{0}}\hat{n}$

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:

$E=\frac{\sigma}{2\epsilon_{0}}\hat{n}+\frac{\sigma}{2\epsilon_{0}}\hat{n} = \frac{\sigma}{\epsilon_{0}}\hat{n}$

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

$E=\frac{\sigma}{2\epsilon_{0}}\hat{n}$

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.

 

[vanhees71]

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.

 

[Demon117]

The electric field at the surface of a conductor is always:

$E=\frac{\sigma}{\epsilon_{0}}\hat{n}$

Each charged face produces an electric field given by:

$E=\frac{\sigma}{2\epsilon_{0}}\hat{n}$

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:

$E=\frac{\sigma}{2\epsilon_{0}}\hat{n}+\frac{\sigma}{2\epsilon_{0}}\hat{n} = \frac{\sigma}{\epsilon_{0}}\hat{n}$

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

If that is the case then wouldn't the Electric field be simply $E=\frac{\sigma}{2\epsilon_{0}}\hat{n}$ for the sheet with surface charge density on one face only?

 

[vanhees71]

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 $z=\pm d/2$ Due to symmetry the potential can only depend on z. Then you have

$$\Delta \Phi(z)=\Phi''(z)=0 \; \Rightarrow \; \Phi(z)=A z$$

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

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

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

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

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

$$A_>=-\frac{\sigma}{\epsilon_0}.$$

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

$$A_<=+\frac{\sigma}{\epsilon_0}.$$

The solution thus reads

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

The electric field is

$$\vec{E}(z)=-\vec{\nabla} \Phi(z)=\begin{cases}<br /> 0 & \text{for} \quad |z|<d/2, \\<br /> \mathrm{sign}(z) \frac{\sigma}{\epsilon_0} \vec{e}_z & \text{for} \quad |z| \geq d/2.<br /> \end{cases}$$

 

[x_engineer]

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.

 

[mitch_1211]

E is always zero inside a conductor.

Provided there is no moving charge of course

 

[Meir Achuz]

Provided there is no moving charge of course

I took the original question to be about electrostatics.

 

[Demon117]

To clarify, this is an electrostatics problem. Thank you all for your support.