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 &amp; \text{for} \quad -d/2&lt;z&lt;d/2 \\<br /> A_&gt; z &amp; \text{for} \quad z \geq d/2 \\<br /> A_&lt; z &amp; \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_&gt;=-\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_&lt;=+\frac{\sigma}{\epsilon_0}.

The solution thus reads

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