# Is electrostatic pressure independent of external field?

I'm reading "Introduction to Eletrodynamics" by David J. Griffiths. And the definition of electrostatic pressure on a conductor makes me confuse.
As Griffiths wrote- because the field inside a conductor is zero, so, according to boundary condition, the field just outside must be $(\sigma/\epsilon_0)\hat{n}$ ,the average $(\sigma/2\epsilon_0)\hat{n}$ and the force per unit area: $$f = \frac{1}{2\epsilon_0}\sigma^2\hat{n}$$ Expressing the pressure in terms of the field just outside the surface: $$P=\frac{\epsilon_0}{2}E^2$$ This pressure tends to draw the conductor into the field.

As I see in the equation, does the Electric field just outside the conductor does not depend on external field? And the Pressure also is independent on the external field. But if external field is, let say, zero, then is there no field for "this pressure" to draw into? Am I missing something? Please enlighten me!

Let me focus on this part of the question:

does the Electric field just outside the conductor does not depend on external field?

In the specific case of the surface of a conductor, the "field just outside the conductor" is the external field. Because the field inside the conductor is zero:

• If you know the surface charge, that determines the field just outside the conductor

• If you know the field just outside the conductor, that determines the surface charge.

This is different from e.g. the field around some fixed arbitrary charges, where the charge produce a local field that has to be added to any external field to get the local total. Because the charges in the conductor can move, the field inside is zero, therefore the entire "external" field has to terminate (or originate) with surface charge.

does the Electric field just outside the conductor does not depend on external field?

I think you just have some misunderstanding of terminology.

The electric field just outside the conductor ($E$ on the diagram below) is part of the external (electrostatic) field. As you know, there is no internal electrostatic field in the conductors. Since the strength of the field may change with the distance from the surface, it is important to point out that the field strength, $E$, used in the formula for the electrostatic pressure is the field strength "just outside the surface".

And the Pressure also is independent on the external field.

The pressure on the surface of the charged conductor is due to the repelling forces between similar charges trying to get away from each other (think about the charged leaves of an electroscope flying apart).

The greater the charge density, the greater the repelling forces between them, the greater the electrostatic pressure. We can also say that the charges, pushed to and distributed over the external surface, are responsible for the external electric field and the strength of that field will also increase with the charge density.

So, both the pressure and the strength of the external electric field increase with or depend on the charge density, but it would not be logical to say that the pressure depends on the external field.

• Thanks V.F. and Bob, I think I get the idea of your answers. The charge density will always auto arrange to make inside's field zero. So any external field, for example from a charge outside the conductor, will make charge density change. So charge density of conductor is exactly the "effect" of all available field. (I missed the point that charge density can change, and this make me fail understanding the logic of equation). – DzungNguyen Jun 27 '18 at 16:24

In steady state, the field just outside the surface of a conductor regardless of whether any external effects exist is normal to the surface of the conductor and proportional to the charge density of the conductor around the point where we are calculating the field.

In steady state is important here as the charge on a conductor will in fact redistribute as any external field is applied to "make" the field inside the conductor zero. As a consequence the net field at a point just outside the surface of a conductor will be dependant only on the surface charge density around that point.

I found this to have the best explanation on electric pressure.