# Average Electric Field over a Spherical Surface

• cwill53
In summary, the conversation discusses the integral that needs to be evaluated and how the associated picture can be obtained using the law of cosines. The character used in the integral is clarified and the electric field is decomposed into parallel and perpendicular components, with the parallel component being proportional to ##\frac{\cos\alpha}{r^2}##. There is a discrepancy with the direction of the electric field and a different expression for the electric field is suggested for clarity.
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.

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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.

andrewkirk said:
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.

Essentially the problem is this integral and its result:

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

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
kuruman said:
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.

kuruman
kuruman said:
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##.

cwill53 said:
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.

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cwill53
kuruman said:
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.

kuruman

## 1. What is the formula for calculating the average electric field over a spherical surface?

The formula for calculating the average electric field over a spherical surface is Eavg = Q/4πεr², where Q is the total charge enclosed within the surface, ε is the permittivity of the medium, and r is the radius of the spherical surface.

## 2. How is the average electric field over a spherical surface different from the electric field at a specific point on the surface?

The average electric field over a spherical surface takes into account the entire surface and the charge enclosed within it, while the electric field at a specific point only considers the charge at that point. The average electric field gives an overall representation of the electric field over the entire surface, rather than just at one point.

## 3. Can the average electric field over a spherical surface be negative?

Yes, the average electric field over a spherical surface can be negative. This may occur if the charge enclosed within the surface is negative, or if the electric field vectors at different points on the surface cancel each other out.

## 4. How does the distance from the center of the spherical surface affect the average electric field?

The average electric field over a spherical surface is inversely proportional to the square of the distance from the center of the surface. This means that as the distance increases, the average electric field decreases. This relationship is described by the formula Eavg = Q/4πεr².

## 5. Can the average electric field over a spherical surface be used to calculate the electric potential at a point on the surface?

Yes, the average electric field over a spherical surface can be used to calculate the electric potential at a point on the surface. The electric potential at a point on the surface is given by V = kQ/r, where k is the Coulomb constant, Q is the charge enclosed within the surface, and r is the distance from the center of the surface to the point. By rearranging this formula, we can see that V = Eavg * r.

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