# Griffith's EM third ed. pp.68 equ 2.12

1. Jan 19, 2005

### quasar987

Griffith's EM third ed. pp.68 equ 2.12 ....

"In the case of a point charge q at the origiin, the flux of E though a sphere of radius r is

$$\int \vec{E} \cdot d\vec{A} = \int \frac{q}{4\pi\epsilon_{0}r^2}\hat{r} \cdot (r^2sin\theta \ d\theta \ d\phi \ \hat{r})$$"

How the hell is dA equal to that? Multiply dr to it and it's dV in spherical coord. The area of a sphere is 4 pi r^2. Shouldn't dA be 8 pi r dr?

Last edited: Jan 19, 2005
2. Jan 19, 2005

### Galileo

You have to consider a small surface element.
You can see it as rectangle of sides $r d\phi$ and $r\sin \theta d\theta$. So the area element is: $r^2\sin \theta d\theta d\phi$.

You can check that this will give the right area for the sphere:

$$A=\int_0^{2\pi}\;\; \int_0^{\pi}r^2\sin \theta d\theta d\phi=4\pi r^2$$

The only way I know how to derive it rigorously is by using calculus:
Parametrize the sphere as:
$$\vec r(\theta,\phi)=\sin \theta \cos \phi \vec i +\sin \theta \sin \phi \vec j + \cos \theta \vec k \qquad 0 \leq \theta \leq \pi , \quad 0\leq \phi \leq 2\pi$$

And calculate $$dA=|\vec r_{\theta} \times \vec r_{\phi}|d\theta d\phi$$ (the Jacobian of the transformation). That'll give the right answer.

I`m sure there are more clever ways to do it though.

Last edited: Jan 20, 2005
3. Jan 19, 2005

### quasar987

Mmh, ok I see it now.

It is unfortunate that both the surface element ($r^2\sin \theta d\theta d\phi$) of a sphere and the infinitesimal increase in area of the sphere ($8\pi r dr$) have the same notation.

4. Jan 20, 2005

### cepheid

Staff Emeritus
^^What is that? Besides, it is never mentioned in Griffiths, so how is it unfortunate?

5. Jan 20, 2005

### dextercioby

I remember seing it pretty easily through the first method prescribed by Galileo...Geometry in this case does more than calculus...Sure,for the volume element,i'd still reccomend the jacobian...

Daniel.

P.S.That notation with "hats" to designate unit vectors sure got me fooled... :grumpy:

6. Jan 20, 2005

### quasar987

What I had in mind is that... The area of a sphere is $A = 4\pi r^2$. So if you differentiate A with respect to r, you get the rate of chage of the area with respect to radius: $\frac{dA}{dr}=8\pi r$. Or, multiplying both sides by dr, you get a differential equation, or function (in the sense that the differential dr is a variable): $dA = 8\pi r dr$. This is saying that for a small increment of radius $\Delta r$, the area increases approximately according to the equation $\Delta A = 8 \pi r \Delta r$... and the smaller the $\Delta r$, the more precise the approximation. Or, we could say that for a $\Delta r$ infinitely small (i.e. infinitesimal) it is no longer an approximation.

It is unfortunate because it got me confused! dA of a surface could either be understood as the dA above (increment in area), or as an element area: $dA = dxdy$. This is why, I think, it is important to write the surface integrals as double integrals $\iint f dA$ (something Mr. Griffiths neglects), so that we don't confuse the "surface element" dA with the "area increment" dA.

7. Jan 21, 2005

### dextercioby

What do i get the feeling that u and only u may mistankenly take the area element as a differential area increment...?? :tongue2: The notation is misleading,i admit,i always denote the former by $dS$ or even $d\vec{S}$ (orientable surfaces are common in physics),while the latter is simply $dA$.

It's the book's fault,but not totally...

Daniel.

8. Jan 21, 2005

### quasar987

Of course, I was tired and my thoughts were foggy... but bookwritters should take this possible state of their readers into account!