Direction of electric field vector on the surface of charged conductor

Consider a negatively charged spherical conductor. On the surface of it, what is the direction of its electric field? Well, the definition of the direction of an electric field is the direction a positive test charge would go if placed at that point. But... it wouldn't move anywhere! So is the electric field here zero? Probably not, but I can't find any information about it online.

Also, what if we considered a positively charged spherical conductor to start with? I mean, according to Newton's Shell Theorem, any particle outside the surface feels a force as if all of the charge were concentrated at its centre. But I'm not convinced that the particle is truly "outside the surface." Rather, it's "on" the surface!

laser said:

Homework Statement: Direction of electric field vector on the surface of charged conductor
Relevant Equations: F=ma

Is F = ma relevant?

laser said:

Consider a negatively charged spherical conductor. On the surface of it, what is the direction of its electric field? Well, the definition of the direction of an electric field is the direction a positive test charge would go if placed at that point.

A better definition (of the direction of electric field) is that it is the direction of the electric force experienced by a positive test charge - radially inwards in this case.

It's analogous to a gravitational field. The gravitational field where you are acts radially inwards towards the centre of the earth. But other forces stop you from accelerating in that direction (hopefully!). Remove the other forces (e.g. the floor collapses) and radially inwards you go!

laser said:

Also, what if we considered a positively charged spherical conductor to start with? I mean, according to Newton's Shell Theorem, any particle outside the surface feels a force as if all of the charge were concentrated at its centre. But I'm not convinced that the particle is truly "outside the surface." Rather, it's "on" the surface!

It doesn't matter if the particle is on the surface or outside the surface. The positive test charge will exprience a radially outwards electric force identical to the force it would experience from a central point charge (of the same size as the surface charge).

BTW, Newton's shell theorem (being pedantic) applies to gravitational fields. But the concept can also be applied to electric fields.

Edit - typo' corrected.

Steve4Physics said:

Is F = ma relevant?

Should've put N/A! :)

Steve4Physics said:

It's analogous to a gravitational field. The gravitational field where you are acts radially inwards towards the centre of the earth. But other forces stop you from accelerating in that direction (hopefully!). Remove the other forces (e.g. the floor collapses) and radially inwards you go!

This makes a lot of sense.

Steve4Physics said:

It doesn't matter of the particle is on the surface or outside the surface. The positive test charge will exprience a radially outwards electric force identical to the force it would experience from a central point charge (of the same size as the surface charge).

Well, if the particle is on the surface... I'm not so sure about that. Because imagine if the test charge was barely inside the surface - according to newton's shell theorem (idk what else to call it :D), there is no force on that test charge. Do you agree?

Steve4Physics said:

It's analogous to a gravitational field. The gravitational field where you are acts radially inwards towards the centre of the earth. But other forces stop you from accelerating in that direction (hopefully!). Remove the other forces (e.g. the floor collapses) and radially inwards you go!

Actually, I think there's a difference. In a gravitational field, there is still mass inside the "gaussian surface". In this case, there is no charge inside the gaussian surface if we are just inside the conductor. "exactly on the surface" though I'm not sure.

laser said:

Well, if the particle is on the surface... I'm not so sure about that.

Remember, you are talking about ideal abstractions. This is partly for convenience and partly because it works accurately enough to apply to real-world systems.

For example, the ideal conductor is a continuous medium with a mathematical discontinuity (the surface) between the inside and outside.

But a real conductor is made from discrete distributed particles with a fuzzy (quantum mechanical) boundary between the inside and the outside.

View the shell theorem as idealised but still providing a very practical, accurate description of real world behaviour. Don't overthink it!

laser said:

Consider a negatively charged spherical conductor. On the surface of it, what is the direction of its electric field?

You are asking about the direction of the electric field and you end up wondering whether it is zero. The direction of the electric field at the surface has direction perpendicular to the surface of the conductor. Here is a simple argument.

If the electric field had a component tangent to the surface, the electrons inside the conductor will move in response to that tangential force and keep on moving until they have no reason to move any more, i.e. they have rearranged themselves to cancel the tangential component. This leaves only the normal component of electric field. Similar reasoning can be used to argue that a static electric field does not penetrate the skin of the conductor.

Steve4Physics said:

But a real conductor is made from discrete distributed particles with a fuzzy (quantum mechanical) boundary between the inside and the outside.

Ah, I see. So it depends on "where" on the surface it is placed. If it's on the outer, it experiences an electric field, if it's on the inner, it doesn't. If it's exactly on the surface... well... we don't talk about that as it's almost impossible to do that?

kuruman said:

You are asking about the direction of the electric field and you end up wondering whether it is zero. The electric field at the surface has direction perpendicular to the surface of the conductor.

I think that is only true for the electric field just outside the surface. Outside the surface, it is normal. Inside the surface, it doesn't exist. What about exactly on the surface?

laser said:

I think that is only true for the electric field just outside the surface. Outside the surface, it is normal. Inside the surface, it doesn't exist. What about exactly on the surface?

In the classical model, charge in a conductor is viewed as a continuous fluid. On the surface of a conductor all one can say is that the electric field is discontinuous across the surface because there is an infinitely thin charge distribution on the surface. The classical model has its limitations. Asking for the value of the electric field inside an infinitely thin film is like asking how many angels can dance on the head of a pin.

If you want to get a better idea of what is going on extremely close to the vacuum-conductor interface, you have to abandon the classical model and consider the "granularity" of the atoms at the surface and the wave nature of the electrons in the conductor. In other words, you need to do a quantum mechanical calculation of sorts.

laser said:

Ah, I see. So it depends on "where" on the surface it is placed. If it's on the outer, it experiences an electric field, if it's on the inner, it doesn't. If it's exactly on the surface... well... we don't talk about that as it's almost impossible to do that?

laser said:

I think that is only true for the electric field just outside the surface. Outside the surface, it is normal. Inside the surface, it doesn't exist. What about exactly on the surface?

Try this...

Consider a charged conducting object as having its free charge in an extremely thin layer, from the object's surface to depth d. That's a very crude model of course but might help.

At the outer surface , the electric field has some value, ##E_o##.
The field smoothly drops from ##E_o## at the surface to 0 at depth d.

I would guess that, in this crude model, the charge layer is typically a few atoms thick, i.e. d~##10^{-9}##m-ish.

An accurate model needs quantum mechanics, as noted by @kuruman.

Edit - typo's