Average Electric Field over a Spherical Surface

[cwill53]

Homework Statement

Griffiths' Electrodynamics, 4th Edition, Problem 3.4

Show that the average electric field over a spherical surface, due to charges outside the sphere, is the same as the field at the center.

Relevant Equations

$$\mathbf{E}_{\textup{average}}=\frac{1}{4\pi R^2}\oint_{\textup{sphere}}^{}\mathbf{E}\, dS$$

The picture above shows the integral that needs to be evaluated, and the associated picture $\cos\alpha$ can be obtained via the law of cosines. I'm simply confused as to where the $\cos\alpha$ comes from in the first place. I just don't see why $\cos\alpha$ is necessary in this expression.

 

[andrewkirk]

I don't recognise the character you have written as the square in the denominator of your integrand. I'll assume it is $r$. That represents the distance from the little patch $dS$ on the sphere's surface to the charge $q$.

Consider the electrical field in the little patch $dS$ due to $q$. the strength of that field is proportional to $\frac1{r^2}$. We can decompose that electrical field into two components, parallel and perpendicular to the $z$ axis, and we get those by multiplying the electric field by $\cos\alpha$ and $\sin\alpha$ respectively. We can ignore the perpendicular component because it exactly cancels out against the corresponding component for the patch we get by rotating our current patch by 180 degrees around the z axis - ie the patch with the same latitude and opposite longitude. Hence we are left with the parallel component, which is proportional to $\frac{\cos\alpha}{r^2}$, as shown in the above integral.

 

[cwill53]

So for a sanity check I decided to use the following approach. Since I don't know how to TeX the fancy r used in Griffiths' book, I'll use $\eta$.

$$\mathbf{E}_{\textup{average}}=\frac{1}{4\pi R^2}\oint_{\textup{sphere}}^{}\mathbf{E}\, dS=\frac{1}{4\pi R^2}\oint_{\textup{sphere}}^{}\left ( \frac{q}{4\pi \epsilon _0\eta ^2} \right )\hat{\mathbf{r}}\, dS$$

$$\frac{1}{4\pi R^2}\frac{q}{4\pi \epsilon _0}\oint_{\textup{sphere}}^{}\frac{\sin\theta \cos\varphi \hat{\mathbf{x}}+\sin\theta \sin\varphi \hat{\mathbf{y}}+\cos\theta \hat{\mathbf{z}}}{\eta^2}dS$$

From the law of cosines, $\eta ^2=R^2+z^2-2Rz\cos\theta$. So

$$\frac{1}{4\pi R^2}\frac{q}{4\pi \epsilon _0}\int_{0}^{2\pi }\cos\varphi \, d\varphi \int_{0}^{\pi }\frac{\sin\theta }{R^2+z^2-2Rz\cos\theta }R^2\sin\theta\, d\theta \hat{\mathbf{x}}=\mathbf{0}$$

$$\frac{1}{4\pi R^2}\frac{q}{4\pi \epsilon _0}\int_{0}^{2\pi }\sin\varphi \, d\varphi \int_{0}^{\pi }\frac{\sin\theta }{R^2+z^2-2Rz\cos\theta }R^2\sin\theta\, d\theta \hat{\mathbf{y}}=\mathbf{0}$$

For the z-component, the integral is

$$\frac{1}{4\pi R^2}\frac{q}{4\pi \epsilon _0}\int_{0}^{2\pi } d\varphi \int_{0}^{\pi }\frac{\cos\theta }{R^2+z^2-2Rz\cos\theta }R^2\sin\theta\, d\theta \hat{\mathbf{z}}$$

But evaluation of the integral

$$\int_{0}^{\pi }\frac{\cos\theta }{R^2+z^2-2Rz\cos\theta }R^2\sin\theta\, d\theta$$

yields this big result (via Symbolab) :

$$-\frac{R}{z}+\frac{R^2\ln \left(\frac{\left|R^2+2Rz+z^2\right|}{\left|R^2-2Rz+z^2\right|}\right)}{4z^2}+\frac{\ln \left(\frac{\left|R^2+2Rz+z^2\right|}{\left|R^2-2Rz+z^2\right|}\right)}{4}$$

$$-\frac{R}{z}+\frac{R^2\left(\ln \left|R^2+2Rz+z^2\right|-\ln \left|R^2-2Rz+z^2\right|\right)}{4z^2}+\frac{\ln \left|R^2+z^2+2Rz\right|-\ln \left|R^2+z^2-2Rz\right|}{4}$$

I don't think this is right, and that makes me wonder if the x and y component integrals are written incorrectly.
Symbolab has been known to make incorrect simplifying assumptions though.

 

[cwill53]

I don't recognise the character you have written as the square in the denominator of your integrand. I'll assume it is $r$. That represents the distance from the little patch $dS$ on the sphere's surface to the charge $q$.

Consider the electrical field in the little patch $dS$ due to $q$. the strength of that field is proportional to $\frac1{r^2}$. We can decompose that electrical field into two components, parallel and perpendicular to the $z$ axis, and we get those by multiplying the electric field by $\cos\alpha$ and $\sin\alpha$ respectively. We can ignore the perpendicular component because it exactly cancels out against the corresponding component for the patch we get by rotating our current patch by 180 degrees around the z axis - ie the patch with the same latitude and opposite longitude. Hence we are left with the parallel component, which is proportional to $\frac{\cos\alpha}{r^2}$, as shown in the above integral.

That makes perfect sense now. Thanks. So I'm guessing now that my approach in my second reply was not correct then.

 

[cwill53]

Essentially the problem is this integral and its result:

$$ \int_{0}^{\pi }\frac{\cos\theta }{R^2+z^2-2Rz\cos\theta }R^2\sin\theta\, d\theta=-\frac{(R^2+z^2)\log\left ( \frac{(R-z)^2}{(R+z)^2} \right )+4Rz}{4z^2}$$

 

[kuruman]

The direction of the electric field as shown in post #3 is normal to the sphere but should not be so. The curly $r$ that Griffiths uses is the vector from the source to the point of observation. I dislike it because it leads to confusion as is the case here. I prefer to use the more conventional (and transparent) expression for the electric field due to a point charge $q$, $$\mathbf{E}=\frac{kq~(\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^{3}}$$where $\mathbf{r}$ = the position of the point of interest and $\mathbf{r'}$ = the position of the source.

Here the source is a charge at distance $z$ on the $z$-axis and the point of interest is on the sphere. Therefore,$$\begin{align} & \mathbf{r}=R\sin\theta\cos\phi~\mathbf{\hat x}+R\sin\theta\sin\phi~\mathbf{\hat y}+R\cos\theta~\mathbf{\hat z}\nonumber \\&
\mathbf{r'} = z~\mathbf{\hat z}.\nonumber \end{align}$$The component of interest that does not integrate to zero is$$E_z=\frac{kq(R\cos\theta-z)}{\left(R^2-2Rz\cos\theta+z^2\right)^{3/2}}.$$

 

[cwill53]

The direction of the electric field as shown in post #3 is normal to the sphere but should not be so. The curly $r$ that Griffiths uses is the vector from the source to the point of observation. I dislike it because it leads to confusion as is the case here. I prefer to use the more conventional (and transparent) expression for the electric field due to a point charge $q$, $$\mathbf{E}=\frac{kq~(\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^{3}}$$where
$\mathbf{r}$ = the position of the point of interest and $\mathbf{r'}$ = the position of the source.
Here the source is a charge at distance $z$ on the $z$-axis and the point of interest is on the sphere. Therefore,$$\begin{align} & \mathbf{r}=R\sin\theta\cos\phi~\mathbf{\hat x}+R\sin\theta\sin\phi~\mathbf{\hat y}+R\cos\theta~\mathbf{\hat z}\nonumber \\&
\mathbf{r'} = z~\mathbf{\hat z}.\nonumber \end{align}$$The component of interest that does not integrate to zero is$$E_z=\frac{kq(R\cos\theta-z)}{\left(R^2-2Rz\cos\theta+z^2\right)^{3/2}}.$$

I see now that my vector was defined incorrectly. Thanks a lot.

 

[cwill53]

The direction of the electric field as shown in post #3 is normal to the sphere but should not be so. The curly $r$ that Griffiths uses is the vector from the source to the point of observation. I dislike it because it leads to confusion as is the case here. I prefer to use the more conventional (and transparent) expression for the electric field due to a point charge $q$, $$\mathbf{E}=\frac{kq~(\mathbf{r}-\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^{3}}$$where $\mathbf{r}$ = the position of the point of interest and $\mathbf{r'}$ = the position of the source.

Here the source is a charge at distance $z$ on the $z$-axis and the point of interest is on the sphere. Therefore,$$\begin{align} & \mathbf{r}=R\sin\theta\cos\phi~\mathbf{\hat x}+R\sin\theta\sin\phi~\mathbf{\hat y}+R\cos\theta~\mathbf{\hat z}\nonumber \\&
\mathbf{r'} = z~\mathbf{\hat z}.\nonumber \end{align}$$The component of interest that does not integrate to zero is$$E_z=\frac{kq(R\cos\theta-z)}{\left(R^2-2Rz\cos\theta+z^2\right)^{3/2}}.$$

Can you help me understand why exactly we need to be careful with the boundaries? I was able to complete the problem by using another substitution.

In the picture, the z-component integral was (as you said)

$$\frac{1}{4\pi R^2}\frac{q}{4\pi \epsilon _0}\int_{0}^{2\pi }\int_{0}^{\pi }\frac{R\cos\theta -z}{(R^2+z^2-2Rz\cos\theta )^{\frac{3}{2}}}R^2\sin\theta \, d\theta d\varphi \, \hat{\mathbf{z}}$$

The troublesome integral was

$$\int_{0}^{\pi }\frac{R\cos\theta -z}{(R^2+z^2-2Rz\cos\theta )^{\frac{3}{2}}}R^2\sin\theta \, d\theta$$

I used the substituions $u\equiv \cos\theta$ and $t\equiv \sqrt{R^2+z^2-2Rzu}$ to get

$$-\int_{\sqrt{R^2+z^2+2Rz}}^{\sqrt{R^2+z^2-2Rz}}\frac{R(-z^2+R^2-t^2)}{2z^2t^2}dt$$

But we have $\sqrt{R^2+z^2-2Rz}=\pm (R-z)$, and $\sqrt{R^2+z^2+2Rz}=\pm (R+z)$. I'm still confused on why we choose which roots to use as bounds in the cases where $z> R$ and $z< R$.

 

[kuruman]

Can you help me understand why exactly we need to be careful with the boundaries? I was able to complete the problem by using another substitution.

View attachment 317422

In the picture, the z-component integral was (as you said)

$$\frac{1}{4\pi R^2}\frac{q}{4\pi \epsilon _0}\int_{0}^{2\pi }\int_{0}^{\pi }\frac{R\cos\theta -z}{(R^2+z^2-2Rz\cos\theta )^{\frac{3}{2}}}R^2\sin\theta \, d\theta d\varphi \, \hat{\mathbf{z}}$$

The troublesome integral was

$$\int_{0}^{\pi }\frac{R\cos\theta -z}{(R^2+z^2-2Rz\cos\theta )^{\frac{3}{2}}}R^2\sin\theta \, d\theta$$

I used the substituions $u\equiv \cos\theta$ and $t\equiv \sqrt{R^2+z^2-2Rzu}$ to get

$$-\int_{\sqrt{R^2+z^2+2Rz}}^{\sqrt{R^2+z^2-2Rz}}\frac{R(-z^2+R^2-t^2)}{2z^2t^2}dt$$

But we have $\sqrt{R^2+z^2-2Rz}=\pm (R-z)$, and $\sqrt{R^2+z^2+2Rz}=\pm (R+z)$. I'm still confused on why we choose which roots to use as bounds in the cases where $z> R$ and $z< R$.

I'm glad you asked that question. This business in the handwritten solution that you posted about the choice of roots is nonsense. Let $A=\sqrt{R^2+z^2-2Rz}$. Quantity $A$ is a positive number and there are no "ifs" and "buts" about it. If I meant $A$ to be negative, I would have written let $A=-\sqrt{R^2+z^2-2Rz}$. Now it is true that I can also write $A=\sqrt{(R-z)^2}$. Having already decided that $A$ is positive, the only choice is $A=z-R$ because $z>R$ since the charge is outside the sphere.

Similarly, if I define $B=\sqrt{R^2+z^2+2Rz}$, it follows that $B=z+R$.

We went through all this not too long ago in this post
https://www.physicsforums.com/threa...a-volume-charge-density.1046926/#post-6816925.
It's the same idea.

 

[cwill53]

I'm glad you asked that question. This business in the handwritten solution that you posted about the choice of roots is nonsense. Let $A=\sqrt{R^2+z^2-2Rz}$. Quantity $A$ is a positive number and there are no "ifs" and "buts" about it. If I meant $A$ to be negative, I would have written let $A=-\sqrt{R^2+z^2-2Rz}$. Now it is true that I can also write $A=\sqrt{(R-z)^2}$. Having already decided that $A$ is positive, the only choice is $A=z-R$ because $z>R$ since the charge is outside the sphere.

Similarly, if I define ##$B=\sqrt{R^2+z^2+2Rz}$, it follows that $B=z+R$.

We went through all this not too long ago in this post
https://www.physicsforums.com/threa...a-volume-charge-density.1046926/#post-6816925.
It's the same idea.

Right right. I remember that problem we did. I will readjust my notes to this answer then. Thanks a lot; your help has been very valuable.