# Electric field inside a flat sheet of copper

1. Sep 20, 2011

### 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.

Last edited: Sep 20, 2011
2. Sep 20, 2011

### clem

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

3. Sep 20, 2011

### 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:

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

4. Sep 21, 2011

### 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.

5. Sep 22, 2011

### Demon117

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?

6. Sep 23, 2011

### 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

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

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 $\vec{n}=-\vec{e}_z$)

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

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

The electric field is

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

7. Sep 24, 2011

### 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.

8. Oct 2, 2011

### mitch_1211

Provided there is no moving charge of course

9. Oct 2, 2011

### clem

I took the original question to be about electrostatics.

10. Oct 2, 2011

### Demon117

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