Electric Fields inside Conductors & Gauss' Law

[pandabear27]

I am just starting to learn AP Physics C E&M, and I have some questions.

1. If I have a metal conductor sphere (neutral or charged), why is it that the electric field inside is always equal to 0?

2. Also, a question about Gauss's Law.
$$E \cdot A = \frac{Q_{enc}}{\varepsilon_0}$$
Since $Q_{enc}$ only accounts for the charges INSIDE of the Gaussian surface, when we solve for E, then does that only account for the electric field that is created by the enclosed charges, or also charges that could be outside of the Gaussian surface?

I would really appreciate any help. Thank you!!

 

[anuttarasammyak]

Hello.

2. all charges both inside and outside the surface contribute to $\mathbf{E}$.
$$\int_S \mathbf{E}\cdot d\mathbf{S}$$ is proportional to enclosed charges.

 

[pandabear27]

Got it, thank you!!

 

[Ibix]

1. If I have a metal conductor sphere (neutral or charged), why is it that the electric field inside is always equal to 0?

In a conductor, charges can move. What will happen to them if there is an E field inside the conductor?

2. Also, a question about Gauss's Law.
$$E \cdot A = \frac{Q_{enc}}{\varepsilon_0}$$

Gauss' Law doesn't say that - it says $$\oint \vec E\cdot d\vec A=\frac{Q_{\mathrm{enc}}}{\varepsilon_0}$$The integral is important because Gauss is making a statement about the flux through a closed surface, not just some random non-closed surface.

Notice that a field line from a charge inside the surface will cross the surface once, and every field line from the charge will have the same sign for $\vec E\cdot d\vec A$ where it crosses. Thus the charge's contribution to the integral must be non-zero.

However, a field line originating outside the surface will cross twice, once inward and once outward, meaning it will make two opposite-signed contributions to the integral. They don't necessarily cancel exactly, but once you do the integral all these contributions do net to zero. So it's only charges inside the surface that matter for calculating the total flux.

 

[pandabear27]

1. I think if there's an E field inside the conductor, then the electrons would feel a force and start moving. Since electrons repel each other, they will go as far as possible, which is along the surface. How can this prove the $E_{field} = 0$?

2. I see, that makes a lot of sense. Thank you so much!!!

 

[Dale]

1. I think if there's an E field inside the conductor, then the electrons would feel a force and start moving. Since electrons repel each other, they will go as far as possible, which is along the surface. How can this prove the Efield=0?

It is not generally true that the E field in a conductor is 0. It is only true in the electrostatic limit where the current density is zero.

 

[Ibix]

1. I think if there's an E field inside the conductor, then the electrons would feel a force and start moving. Since electrons repel each other, they will go as far as possible, which is along the surface. How can this prove the $E_{field} = 0$?

If they move when there's an E field, when will they stop moving? Might the changed arrangement of the electrons be able to provide that circumstance when added to your external field?

(Note that there might not be such a circumstance, as @Dale commented, but there will be for something like a conducting sphere near a charge.)

 

[Meir Achuz]

We derive Gauss's law from Coulomb's law for a single point charge.
We start with the surface integral,
$$
\oint {\bf dA}\cdot{\bf E}=q \oint{\bf d A}\cdot\frac{{\bf r}}{r^3},
$$
of the normal component of ${\bf E}$ over a closed surface surrounding the point charge.

The vector differential of area, ${\bf dA}$, is an infinitesimal surface element of magnitude dA. Since it is infinitesimal, it approaches a plane surface, tangent to the closed surface. By convention, its vector direction is along the outward normal to the closed surface. The integrand,$\frac{{\bf \hat r}\cdot{\bf dA}}{r^2}$, of the surface integral, can be recognized as the definition of the solid angle, $d\Omega$, subtended by the differential surface.
Then the surface integral can be written as
$$\oint{\bf dA}\cdot{\bf E}=q\oint d\Omega=4\pi\, q,$$
with the factor $4\pi$ arising as the magnitude of the total solid angle of any closed surface.

If the point charge $q$ were located outside the closed surface, then the surface integral would be zero.
If a plane surface is made to cut the closed integration surface into two parts, then the integrated solid angle over each part of the surface will equal in magnitude the solid angle subtended by the part of the plane surface inside the closed surface. But the two solid angles will be of opposite sign and just cancel. Thus the integral over the closed surface will be $4\pi q$ for a point charge inside the surface, and zero for a point charge outside the surface.

For a collection of point charges, only those inside the surface will contribute to the integral, and we have Gauss's law,
$$\oint {\bf dA}\cdot{\bf E}=4\pi\, Q_{\rm enclosed},$$
where $Q_{\rm enclosed}$ is the net charge within the surface.