I Gauss' Law - Does it apply even when there are external fields?

[randomuser1]

TL;DR Summary

Hi all. I have a question regarding Gauss law. I have learnt that Gauss law applies irrespective of external charges. However, there is a situation which I can seem to reconcile. Consider the case of a coaxial cable, with two concentric conductors separated by a layer of air. The electric field strength in the air region can be found by considering a Gaussian surface which is cylindrical of radius r. It will enclose the inner wire which has say charge per unit length of +λC/m. However, what if

Hi all. I have a question regarding Gauss law. I have learnt that Gauss law applies irrespective of external charges. However, there is a situation which I can seem to reconcile. Consider the case of a coaxial cable, with two concentric conductors separated by a layer of air. The electric field strength in the air region can be found by considering a Gaussian surface which is cylindrical of radius r. It will enclose the inner wire which has say charge per unit length of +λC/m. However, what if the charge per unit length on the outer conductor is also +λC/m? By Gauss law, the Gaussian surface still encloses charge and therefore there is still an electric field. However, I would expect electric field of 0 between two surfaces with the same charge density?

Thanks!

 

[BvU]

Hi, !

On the inside of the outer conductor there will be an induced charge of $-\lambda$ C/m.
On the outside of the outer conductor there will be a charge of $+2\lambda$ C/m, exactly in agreement with Gauss' law ( so outside the cable an electric field as from a line charge with $+2\lambda$ C/m !

[edit]That way there is zero total charge inside a cylindrical gaussian surface between the inside and the outside of the outer conductor, so zero electric field between the inside and the outside of the outer conductor -- exactly as it should be for a conductor.

[note] in everyday practice the outer conductor is grounded and the induced charge on the inside shields the charge on the inner cable, and outside the outer conductor the electric field is zero

$\$

 

[kuruman]

However, I would expect electric field of 0 between two surfaces with the same charge density?

Why? How do you think the charge is distributed on the outer conductor? Before you answer, make sure that what you say is consistent with . . . Gauss's law.

 

[PeroK]

However, I would expect electric field of 0 between two surfaces with the same charge density?

Why would you expect this? By the shell theorem, it's clearly not true for concentric spherical surfaces.

 

[randomuser1]

Why would you expect this? By the shell theorem, it's clearly not true for concentric spherical surfaces.

Hi, thanks for the reply. I’m not exactly familiar with the shell theorem for electrostatics, is it that a spherical shell would behave exactly like a point charge with the same charge (outside of the shell)? How would this idea address my question? Thanks.

 

[PeroK]

Hi, thanks for the reply. I’m not exactly familiar with the shell theorem for electrostatics, is it that a spherical shell would behave exactly like a point charge with the same charge (outside of the shell)? How would this idea address my question? Thanks.

There are two parts to the shell theorem:

Outside a uniformly charged spherical shell, the electric field is the same as though all the charge were concentrated at the centre of the sphere.

Inside a uniformly charged spherical shell, the electric field is zero.

You can extend these to the cases of spherically symmetrically charged spheres and partially hollow spheres.

The easiest proof is to use Gauss's theorem! It's a good exercise to prove it using the electric field of a point charge and calculus. It's easier using electric potential. It's essentially the same proof for Newtonian gravity.

 

[randomuser1]

There are two parts to the shell theorem:

Outside a uniformly charged spherical shell, the electric field is the same as though all the charge were concentrated at the centre of the sphere.

Inside a uniformly charged spherical shell, the electric field is zero.

You can extend these to the cases of spherically symmetrically charged spheres and partially hollow spheres.

The easiest proof is to use Gauss's theorem! It's a good exercise to prove it using the electric field of a point charge and calculus. It's easier using electric potential. It's essentially the same proof for Newtonian gravity.

Hi PeroK, thanks for the reply. "Inside a uniformly charged spherical shell, the electric field is zero." For the case of concentric spherical surfaces, this does make sense for the inner sphere; the electric field within the inner sphere is 0. However, does this argument extend for the outer sphere? Because there would be electric field between the inner and outer sphere right?

Thanks in advance. Cheers!

 

[PeroK]

Hi PeroK, thanks for the reply. "Inside a uniformly charged spherical shell, the electric field is zero." For the case of concentric spherical surfaces, this does make sense for the inner sphere; the electric field within the inner sphere is 0. However, does this argument extend for the outer sphere? Because there would be electric field between the inner and outer sphere right?

Thanks in advance. Cheers!

Yes, by the principle of superposition, the field between the shells is the field due to the inner shell only.

 

[kuruman]

However, does this argument extend for the outer sphere? Because there would be electric field between the inner and outer sphere right?

In figure (A) on the right is shown a conducting sphere with an irregular cavity inside it. Total charge $Q$ is placed on the sphere. Where does it go?

Answer: Consider two Gaussian surfaces indicated by dashed lines, blue enclosing the sphere and red enclosing the cavity. Since the electric field is zero everywhere inside the conductor, it is also zero everywhere on the red surface. By Gauss's law the total charge enclosed by the red surface must be zero. It follows that charge $Q$ can only be on the surface of the sphere. If not, you should be able to find a red Gaussian surface that will enclose some charge which will make the field inside the conductor non-zero.
The blue Gaussian surface encloses total charge $Q$. Because the spherical surface symmetry, it will be distributed uniformly over the surface and the electric field outside the sphere at distance $r$ from the center will be the same as if total charge $Q$ were concentrated at the center of the sphere, $$E=\frac{Q}{4\pi\epsilon_0 r^2}.$$Of course, the field is not the same everywhere on the blue surface, but at any point on it, it will be given by the above expression.

Now suppose I introduce additional charge $-q$ at some point inside the cavity (see figure on the right). How does that change things?
Answer: The field inside the conductor is still zero which does not change the finding that the red Gaussian surface must enclose zero net charge. For that to happen, total charge $+q$ must be on the inner surface as shown. Since the total charge on the sphere has not changed, the charge on the surface of the sphere must now be $Q-q$. This means that the electric field outside the sphere is now given by $$E=\frac{(Q-q)}{4\pi\epsilon_0 r^2}.$$Also note that in case (A) the field inside the cavity is zero. In case (B) it is not. There is an electric field inside the cavity with field lines starting at the positive charges on the inside surface and ending at the negative charge $-q.$ If you move $-q$ in the cavity to a new position, you will change the electric field inside and the charge distribution on the inner surface but not the total. Since the electric field is always zero inside the conductor, there is no way for the inside to communicate the the outside that the charge has moved. Someone monitoring the electric field outside the sphere will be unable to tell whether the charge inside is being moved or not.

 

[A.T.]

Hi PeroK, thanks for the reply. "Inside a uniformly charged spherical shell, the electric field is zero."

Maybe it's clearer stated as:

"Inside a uniformly charged spherical shell, the electric field caused by the charge on that shell is zero."