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## Homework Statement

Use Gauss' Law to show that all static charge on a conductor lies entirely on the surface of the conductor.

## Homework Equations

at a point outside the shell, this would be the final formula supposedly. E(r)= [itex]\frac{Q_s}{4\pi r^2 \epsilon}[/itex]

## The Attempt at a Solution

Sorry if this is in the wrong section, but its technically a homework question.

The proof of this goes like:

From Gauss' law of Gravitation, we know all gravitational force acts on the outside of a spherical shell, because all the mass lies on the outside of the shell.

If we apply Gauss' law of gravitation to a spherical shell, we get:

[itex]\Gamma(4\pi r^2)[/itex] = M

_{s}

[itex]\rightarrow[/itex] [itex]\Gamma (r) [/itex] = -[itex]\frac{M_s}{4\pi r^2}[/itex]

Thus, using the gravitational flux density formula ( g(r) = 4[itex] \pi[/itex]r

^{2})

We find that the gravitational field strength at point P is :

g(r)= -[itex]\frac{G M_s}{ r^2} [/itex][itex]\widehat{r}[/itex]

Since all mass is contained on the outside, if we apply Gauss' law to any point on the inside of the shell, we get a field strength of 0, as so.

[itex]\Gamma (4\pi r^2) [/itex] = 0

[itex]\Gamma (r) [/itex] = 0

[itex]\rightarrow[/itex] g(r) =0

If we are to apply the following transformations to the above, this can be written in terms of electric charge.:

g(r)[itex]\Rightarrow[/itex] E(r)

G[itex]\Rightarrow[/itex] [itex]\frac{1}{4\pi \epsilon}[/itex]

M

_{s}[itex]\Rightarrow[/itex] Q

_{s}

This gives us:

E(r)[itex]\frac{Q_s}{4\pi \epsilon r^2}[/itex] for a point on the surface of the shell

and

E(r)= 0 for a point inside the shell.

Apparently that is the solution, but it seems very.... lazy I guess Is that enough info on it? Or am i missing a big problem? Or can I sign Q.E.D and be done with it? :tongue:

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