I have a couple clarifying questions:
(A) in $E=kq/r^2$ for the area between a sphere inside another spherical shell, the sphere inside is considered a point charge. However, the sphere inside has a radius of say $.1 \text{ m}$, if I wanted to find the E-field at the radius value of the inner sphere, would i plug in .1 for $r$? Isn't $r$ supposed to be zero since it's the distance from a point charge to a point in space? Or is this r just the radius of the gaussian sphere around the inner sphere?
(B) Say the inner sphere has a charge of -1 and the outer spherical shell has a net charge of +2. Then I connect the two conductors with a wire, because of conservation of charge, will the charge on the outer surface of the outer spherical shell be still +1 with an electric field of zero inside the whole spherical shell. i.e there will be a charge of zero on the inner sphere and a charge of zero on the inner surface of the outer spherical shell?
(C) Is the work done by the electric field $= -q\Delta V$ or just $q\Delta V$? Is work done by the field always equal to the negative change in potential energy? Does the $\Delta V$ already account for the negative?
Best Answer
(A) When using Gauss' law for spherically symmetric systems, $r$ is evaluated at the radius of the Gaussian surface. I understand where the confusion comes from, because in the "standard" equation for the electric field, the distance $r$ is defined as the distance between the point where you're measuring the field and the point where the charge is located, and in the case of a spherical surface, you might think that measuring the field on the surface that since you are infinitesimally close to the surface, $r$ should be zero. In a real system, if you were able to get that close to the surface, you would actually have to start worrying about where the quantized charges are located, so things would get complicated close to the surface. For these types of problems, it is assumed that the charges are "smoothed" out over the surface, and using Gauss' law comes in handy. Gauss' Law is used in the following way.
$\oint \vec{E}\cdot d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$
Since you chose a spherical surface $\vec{E}$ and $d\vec{a}$ are parallel. Also, $\vec{E}$ is constant over the surface, so the integral is just the following:
$E\oint da = \frac{Q_{enc}}{\epsilon_0}$
where $\oint da= 4\pi r^2$, and $r$ is the radius of the Gaussian surface. So,
$\vec{E} = \frac{Q_{enc}}{4\pi r^2} \hat{r}$
Now, you can see that $r$ is the radius of the sphere, and the only place that the size of the charged sphere would matter is when determining whether the charge on the surface of the sphere is enclosed or not. Since on a conductor, all of the charge is on the outer surface, at the surface itself, the charge is generally not considered to be enclosed, but just above the surface (infinitesimally), the charge is considered to be enclosed (all of the charge on the sphere).
(B) In this case, you have to remember that conductors do not like to have electric fields on the inside. So when connecting the two sphere's by a conduction wire, any net charge will want to move to the outer surface. So you are right in this case, the charge on the inner sphere is zero and the net charge is on the outer sphere.
(C) The work done by an electric field on a particle is $q\Delta V$. The way to think of it is that the work done on something is equal to the change in potential energy of that object. In this case, the change in the potential energy of a charge is given by $\Delta U = q\Delta V$. Now, the sign of $q$ and $\Delta V$ should be determined by the system.
For instance lets consider free charges at rest in an electric field. A proton has $q=+e$ and an electron has $q=-e$. Protons move from high potential to low potential, and electrons move from low potential to high potential. So, with $\Delta V = V_f - V_i$, in both cases you see a negative work done by an electric field, because the charges will move toward a position that lowers their potential energy. In order to do (positive) work on a charge, you need to raise the potential energy. This is something that batteries do.