Parameterize Radial Vector of Electric Field due to Spherical Shell

[Cedric Chia]

Homework statement:
Find the electric field a distance z from the center of a spherical shell of radius R that carries a uniform charge density σ.

Relevant Equations: Gauss' Law
$$\vec{E}=k\int\frac{\sigma}{r^2}\hat{r}da$$

My Attempt:
By using the spherical symmetry, it is fairly obvious that the horizontal component of electric field at z cancel, leaving only $$\vec{E}=E_{z}\hat{z}$$
and from there we can get rid of the complexity of the $$\hat{r}$$
and continue... BUT !
What if I'm not aware of this symmetry property of the spherical shell and must do the parameterization? I know the x component and y component of this vector: $$\vec{r}=<-Rsin\phi cos\theta ,-Rsin\phi sin\theta, zcos\phi>$$
is correctly parameterized in this way but the z component doesn't seem right...
i.e. when:$$\theta=0$$ $$\phi=180$$
the z component should be:
$$R+\text{distance from origin to the point outside the sphere}$$
in order to get from the bottom of the sphere to the top and some extra distance to get to the point
What is the z component in this parameterization? Please HELP

 

[kuruman]

The usual expression for the electric field due to charge $Q$ at the origin is $$\vec E=\frac{kQ}{r^2}\hat r$$ The Cartesian components of the radial unit vector are given by $\hat r=\sin\theta\cos\phi~\hat x+\sin\theta\sin\phi~\hat y+\cos\theta~\hat z$ and $r^2=x^2+y^2+z^2$. When $\theta =0$, $\phi$ is irrelevant because the point of interest is on the z-axis. The electric field can only have a z-component whether you recognize the symmetry or not. In this special case, $\hat r =\hat z$, $r = z$ and you can write $$\vec E=\frac{kQ}{z^2}\hat z.$$Does that answer your question?

 

[Cedric Chia]

The usual expression for the electric field due to charge $Q$ at the origin is $$\vec E=\frac{kQ}{r^2}\hat r$$ The Cartesian components of the radial unit vector are given by $\hat r=\sin\theta\cos\phi~\hat x+\sin\theta\sin\phi~\hat y+\cos\theta~\hat z$ and $r^2=x^2+y^2+z^2$. When $\theta =0$, $\phi$ is irrelevant because the point of interest is on the z-axis. The electric field can only have a z-component whether you recognize the symmetry or not. In this special case, $\hat r =\hat z$, $r = z$ and you can write $$\vec E=\frac{kQ}{z^2}\hat z.$$Does that answer your question?

Thank you for your help but I'm actually looking for an alternative solution.
For example:
When finding electric field distance z perpendicular above the mid point of a line segment, we usually parameterize:
$$\vec{r}=<-x,0,z>$$
in order to get from any point on the line segment to the point on the z axis
When finding electric field distance z perpendicular above the mid point of a ring with radius R, we use:
$$\vec{r}=<-Rcos\phi ,-Rsin\phi , z>$$
so that any point on the ring can get to the point above the mid point
Now back to my question: a spherical shell, what parameterization should we use in order to get from surface of the sphere to the point which lies on the z axis?
Thank you again.

 

[kuruman]

Ah, I think I understand. Your $\vec r$ is a position vector from an element of charge to a point of interest on the z-axis. (If you have the EM textbook by Griffiths, it is his "curly-r"). It is the difference $(\vec r - \vec r')$ where $\vec r$ is the position vector of the point of interest and $\vec r'$ is the position vector of the element of charge. To be clear about what's what, primed coordinates refer to position in the charge distribution and unprimed elements refer to field coordinates. All integrals are done with respect to primed coordinates.

For the ring that you mentioned,
$\vec r = z~\hat z$ and $\vec r'=R \cos\phi' ~\hat x+R \sin\phi' ~\hat y$ so that
$(\vec r - \vec r')=-R \cos\phi' ~\hat x-R \sin\phi' ~\hat y+z~\hat z$
which agrees with what you have.

For the shell,
$\vec r'=R( \sin\theta' \cos\phi' ~\hat x+ \sin\theta'\sin\phi' ~\hat y+ \cos\theta'~\hat z).$
I leave it up to you to find $(\vec r - \vec r')$. To avoid future confusion, I recommend that you stop using $\vec r$ when you mean to write the difference $(\vec r - \vec r')$.

 

[Cedric Chia]

Ah, I think I understand. Your $\vec r$ is a position vector from an element of charge to a point of interest on the z-axis. (If you have the EM textbook by Griffiths, it is his "curly-r"). It is the difference $(\vec r - \vec r')$ where $\vec r$ is the position vector of the point of interest and $\vec r'$ is the position vector of the element of charge. To be clear about what's what, primed coordinates refer to position in the charge distribution and unprimed elements refer to field coordinates. All integrals are done with respect to primed coordinates.

For the ring that you mentioned,
$\vec r = z~\hat z$ and $\vec r'=R \cos\phi' ~\hat x+R \sin\phi' ~\hat y$ so that
$(\vec r - \vec r')=-R \cos\phi' ~\hat x-R \sin\phi' ~\hat y+z~\hat z$
which agrees with what you have.

For the shell,
$\vec r'=R( \sin\theta' \cos\phi' ~\hat x+ \sin\theta'\sin\phi' ~\hat y+ \cos\theta'~\hat z).$
I leave it up to you to find $(\vec r - \vec r')$. To avoid future confusion, I recommend that you stop using $\vec r$ when you mean to write the difference $(\vec r - \vec r')$.

Thank you! You really made my day! I'm using the Griffiths Textbook but never thought of using vector addition on "curly-r / script-r" $$\text{script r} = \vec{r}-\vec{r'}$$
where $$\text{script r}:=\text{the vector starts at source charges (the sphere) and ends at point z}$$
$$\vec{r}:=\text{the vector starts at origin and ends at point z}$$
$$\vec{r'}:=\text{the vector starts at origin and ends at source charges (the sphere)}$$
Thanks again!

 

[vanhees71]

Since you have radial symmetry here, of course you should work in spherical coordinates. Due to this symmetry the electric field must be of the form $\vec{E}(\vec{r})=E_r(r) \vec{e}_r$. Now use this in Gauss's Law in integral form,
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} Q_V,$$
where $V$ is an arbitary volume, $\partial V$ its boundary surface, and $Q_V$ the charge inside the volume.

In your case you choose of course a sphere with radius $r$ around the origin for $V$. Then you get uniquely $E_r(r)$ everywhere.

Another more simple way is to use the electric potential, which is a function of $r$ only due to the spherical symmetry. Then you only need
$$\Delta \phi=\frac{1}{r} \partial_r^2 (r \phi)=\frac{1}{\epsilon_0} \rho,$$
where in your case
$$\rho(\vec{r})=\rho(r)=\frac{1}{4 \pi a^2} \delta(r-a).$$