Find the electric field inside and outside of a spherical shell superposition

[Davidllerenav]

Homework Statement

Given a spherical shell of radius R and charge density $\sigma_0$, find the electric field at point A r<R and point B r>R using the superposition principle. With this results, find the analogous problem with a solid sphere with charge density $\rho_0$

Relevant Equations

$\vec E=\frac{1}{4\pi\epsilon_0}\int \frac{\sigma}{r^2}\hat{r}$

Hi! I need help with this problem. I tried to solve it by saying that it would be the same as the field of a the spherical shell alone plus the field of a point charge -q at A or B. For the field of the spherical shell I got $E_1=\frac{q}{a\pi\epsilon_0 R^2}=\frac{\sigma}{\epsilon_0}$ and for the point charge $E_2=\frac{-q}{4\pi\epsilon_0 r^2}$, I said that -q was the same as q, and so I could write it as $E_2=\frac{-\sigma R^2}{\epsilon_0 r^2}$. After thaht I add them and I got $E=\frac{\sigma}{\epsilon_0}[1-\frac{R^2}{r^2}]$. As I understand, I was meant to get $E=0$, since at A r<R. what am I doing wrong?

 

[BvU]

find the electric field at point A r<R and point B r>R using the superposition principle

If that is the litteral rendering of the problem statement, you can't pretend the field of the shell is known, as in

saying that it would be the same as the field of a the spherical shell alone plus the field of a point charge -q at A or B

Morerover, you can't add a charge -q, because it isn't there !

I read the exercise as: add contributions from smartly chosen parts of the shell to find E for r<R and for r>R

Then in part b, the solid sphere is a sum of shells

 

[Davidllerenav]

If that is the litteral rendering of the problem statement, you can't pretend the field of the shell is known, as in
Morerover, you can't add a charge -q, because it isn't there !

I read the exercise as: add contributions from smartly chosen parts of the shell to find E for r<R and for r>R

Then in part b, the solid sphere is a sum of shells

I see, I tried to do it by adding the -q charge because I tought I could solve it as a ploblem of a sphere with a cavity.

How should I apply the superposition principle? I was thinking of first using the charges $dq$ on the same line as the point, since those two would produce a field straight to A. all the other little charges $dq=\sigma da$ would cance in pairs, such that only the horizontal component survives. Am I correct?

 

[kuruman]

I would do it formally by starting with the general expression$$\vec E=\frac{1}{4\pi\epsilon_0}\int \frac{(\vec r-\vec r')}{|\vec r-\vec r'|^3}\sigma~ dA'$$Without loss of generality, you can pick the observation point at $\vec r = z ~\hat k$ but the source point must be general, $\vec r'=R(\sin \theta '\cos\phi'~\hat i+\sin \theta '\sin\phi'~\hat j+\cos \theta '~\hat k).$ Then you have to do three integrals, one for each component of the electric field. You should verify that the $x$ and $y$ components vanish as expected.

 

[BvU]

Am I correct?

No. Only the non-axial components cancel.

[edit] sorry, forgot to post reply this afternoon.

Kuruman's advice is good. What is the locus of points with the same $|\vec r-\vec r'|$ ?

 

[Davidllerenav]

I would do it formally by starting with the general expression$$\vec E=\frac{1}{4\pi\epsilon_0}\int \frac{(\vec r-\vec r')}{|\vec r-\vec r'|^3}\sigma~ dA'$$Without loss of generality, you can pick the observation point at $\vec r = z ~\hat k$ but the source point must be general, $\vec r'=R(\sin \theta '\cos\phi'~\hat i+\sin \theta '\sin\phi'~\hat j+\cos \theta '~\hat k).$ Then you have to do three integrals, one for each component of the electric field. You should verify that the $x$ and $y$ components vanish as expected.

Ok, I'll try it. I have one question, is it necessary to do it with vectors? If I know that the field should be in the radial direction, can I work with magnitudes o not?

 

[Davidllerenav]

No. Only the non-axial components cancel.

[edit] sorry, forgot to post reply this afternoon.

Kuruman's advice is good. What is the locus of points with the same $|\vec r-\vec r'|$ ?

All points with the same vector $|\vec r-\vec r'|$ are those on the surface of the shell, right?

 

[kuruman]

All points with the same vector $|\vec r-\vec r'|$ are those on the surface of the shell, right?

All the source points are on the surface of the shell. How would you describe the locus of all points on the shell that have the same $|\vec r-\vec r'|$?

 

[Davidllerenav]

All the source points are on the surface of the shell. How would you describe the locus of all points on the shell that have the same $|\vec r-\vec r'|$?

I don't know what locus means, but I think that all points on the shell that have the same distance to the point would form a circle of radius $|\vec r-\vec r'|$.

 

[kuruman]

A "locus" is the set of all points that share a common property. Example: A circle is the locus of all points that are equidistant from a single point in a two-dimensional plane. Saying they form a circle of radius $|\vec r-\vec r'|$ is not specific enough. How is this circle oriented? Remember that all you have is a shell and a point P outside (or inside) the shell. Given only these two items, how will you draw this circle on the shell?

 

[Davidllerenav]

A "locus" is the set of all points that share a common property. Example: A circle is the locus of all points that are equidistant from a single point in a two-dimensional plane. Saying they form a circle of radius $|\vec r-\vec r'|$ is not specific enough. How is this circle oriented? Remember that all you have is a shell and a point P outside (or inside) the shell. Given only these two items, how will you draw this circle on the shell?

By oriented, do you mean clockwise or counterclockwise? I don't know.

 

[BvU]

When in doubt, make a sketch to clear up your thinking !

 

[kuruman]

By oriented, do you mean clockwise or counterclockwise? I don't know.

Along what direction (positive or negative, it doesn't matter) is the normal to the plane of the circle? The suggestion by @BvU to make a sketch is certainly helpful.

However, you can also forget about the circle of radius $|\vec r-\vec r'|$ and just do the three integrals as I suggested in #4. When you are done with the integral over $\phi '$ (do that first), you will have verified that $E_x=E_y=0$ while the integral for $E_z$ will be the same as the one you get when you consider said circle.