# I Charge on conductors

1. Apr 14, 2016

### Shreyas Samudra

- David J Griffiths

But how s it valid for some conductor like →

2. Apr 14, 2016

### Shreyas Samudra

3. Apr 14, 2016

### Staff: Mentor

Given:
1) Maxwells equations.
2) Electrostatic: $J=0$, $d\rho/dt=0$, $dE/dt=0$, $dB/dt=0$
3) Conductor: $J=\sigma E$

Assume there is charge in the interior. Then by Gauss law there is an E field in the interior. Then by assumption 3) there is a current. This contradicts assumption 2). Therefore there can be no charge in the interior.

4. Apr 14, 2016

### Shreyas Samudra

So does that mean -
if some charge is given to an arbitrarily shaped conductor, it will surface and regardless of everything; the field inside will be zero and the surface will be equipotential

5. Apr 17, 2016

### Shreyas Samudra

somebody !!

6. Apr 17, 2016

### vanhees71

The reasoning is very simple. I don't know, why Griffiths likes to obscure things in his textbooks. In electrostatics by definition we have all fields and sources time-independent and also no current density (see #3). Then within a conductor you have
$$\sigma \vec{E}=\vec{j}=0 \; \Rightarrow \; \vec{E}=0.$$
The electric charge density at every point is given by Gauss's Law (Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{E}=\rho.$$
Since $\vec{E}=0$ inside the conductor also $\rho=\vec{\nabla} \cdot \vec{E}=0$ inside the conductor. So if the conducting body is overall charged, the charge must be on its surface.

Caveat: To be more accurate, of course microscopically the entire body consists of charged particles (atomic nuclei and electrons). So there are charges inside the body but they are compensating each other precisely in the static case, when looked from a macroscopic (coarse-grained) point of view. So within macroscopic electrodynamics the macroscopic charge distribution vanishes in conductors for electrostatic situations.

7. Apr 17, 2016

### Shreyas Samudra

so #4 is right ?

8. Apr 17, 2016

### Shreyas Samudra

So does that mean -
if some charge is given to an arbitrarily shaped conductor, it will surface and regardless of everything; the field inside will be zero and the surface will be equipotential

9. Apr 17, 2016

### Shreyas Samudra

by the way can you put this physically , I don't know this math

10. Apr 17, 2016

### Staff: Mentor

Only in the electrostatic case, yes. This does not hold in magnetostatics or electrodynamics.

11. Apr 19, 2016

### SammyS

Staff Emeritus
Yes. The surface is equipotential.

That does not mean that the surface charge is uniformly distributed.

12. Apr 20, 2016

### Shreyas Samudra

Then we should have current over the surface, as its made of conductor, does that happen in steady state ?
It won't be electrostatics then !

13. Apr 20, 2016

### SammyS

Staff Emeritus
I didn't mention current, nor did I imply it.

The surface charge density is not one uniform value over the whole surface. It varies from location to location, but it is static.

14. Apr 22, 2016

### Shreyas Samudra

that means it is equipotential

15. Apr 22, 2016

Yes.