Calculating Flux through a Sphere using Cylindrical Coordinates

[bowlbase]

I was told it might be better to post this here.

Homework Statement

The trick to this problem is the E field is in cylindrical coordinates.
$E(\vec{r})=Cs^2\hat{s}$

Homework Equations

$\int E \cdot dA$

The Attempt at a Solution

I tried converting the E field into spherical coords and I can find the flux that way but it is a complicated answer. The problem suggests keeping the field in cylindrical and doing the integral of the circle in cylindrical instead of spherical. I'm sort of lost on how I would do that. Would I have the limits of s be 0→R and z -R→R and $\phi$ the same as hat it would normally be?

I doubt it is that simple but since I've never tried to use non-optimal coordinates for an object I'm not entirely sure how I would go about this.

Also, this is the work I've done:

For example

$E=Cs^2\hat{s}$
$s=rsin(\theta)$ and $\hat{s}=sin(\theta)\hat{r}+cos(\theta)\hat{\theta}$
so $E=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})$
$\int E \cdot dA=E4\pi r^2=4\pi r^2(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})$

The next step it asks me to calculate the divergence of E and then graph it on the sz plane.

I can do this with the original equation but I now have answers in two different coordinate systems. Which I suppose sounds fair since they did gave me two also.

$∇\cdot E=\frac{1}{s}\frac{∂}{∂s}(sE_s)$
$=\frac{C}{s}(3s^2)=3Cs$

Finally, it asks that I now do the integral $\int (∇\cdot E) dV$ to show that the two methods are equivalent. At first glance I would say they are not. So I probably made a mistake somewhere.

 

[qbert]

so $E=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})$
$\int E \cdot dA=E4\pi r^2=4\pi r^2(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})$

Now supposing you got E right you need d{\bf A} = r^2 \sin \theta d \theta d\phi \widehat{r},
and that product is the inner product.

 

[bowlbase]

Okay,

$\int_0^{2\pi}\int_0^{\pi}r^2sin^2(\theta)(sin(\theta)\hat{r}+cos( \theta)\hat{\theta})r^2sin(\theta)d\theta d\phi$
=$2\pi r^4\int_0^{\pi}sin^4(\theta)\hat{r}+sin^2(\theta)cos(\theta)\hat{\theta}$

second integral is zero. first integral evaluates to $\frac{3\pi}{8}$

so finally the flux is: $\frac{3}{4}\pi^2r^4$

edit:corrected power on sine

 

[qbert]

I'm not sure what you are doing.

{\bf E}=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})
and
d{\bf A} = r^2 \sin \theta d \theta d\phi \widehat{r},
tell you
{\bf E} \cdot d{\bf A} = r^4 sin^4(\theta) d\theta d\phi.

 

[bowlbase]

I did $sin^4(\theta)$ but put 3 up there instead. What we have then is the same since the theta component goes to zero in the integral.

 

[qbert]

that's what I'm worried about. the right answer for the wrong reason. do you see what's different
between what I'm saying and what you are?

 

[bowlbase]

I see that you are only doing the $\hat{r}$ direction but I don't know why you would ignore the $\hat{\theta}$.

 

[Dick]

I see that you are only doing the $\hat{r}$ direction but I don't know why you would ignore the $\hat{\theta}$.

Because as qbert has been trying to tell you, ${\bf E} \cdot d{\bf A}$ isn't a vector. It's a scalar. It's the inner product of the vector ${\bf E}$ and the vector $d{\bf A}$.

 

[qbert]

There is no \widehat{\theta} direction, it is wiped out when you take the inner product in {\bf E} \cdot d{\bf A}, and that is my point. You have to do the inner product when you're doing flux integrals.
It's only the piece of electric field which pierces the surface element which contributes to flux.

 

[bowlbase]

I'm trying to figure out what you guys mean by inner product. I haven't heard this term before.

 

[Dick]

I'm trying to figure out what you guys mean by inner product. I haven't heard this term before.

Your relevant equation has a dot in it that means 'dot product', same thing as 'inner product'. Do you know what dot product means? Also sometimes called scalar product. It's an operation between two vectors that give you a scalar (i.e. a number).

 

[bowlbase]

I know dot product. I've just never heard it called inner product. However, I don't understand, physically, why we would ignore one direction of the field. It's one thing to say if the direction of the field is only in $\hat{r}$ then that is all fine and understandable (if I did the integral in cylindrical coordinates this would work wonderfully). But, if I had 3 components to the field direction would I still ignore the other two in favor of r? I know they are all perpendicular but I don't grasp why they go away physically.

 

[Dick]

I know dot product. I've just never heard it called inner product. However, I don't understand, physically, why we would ignore one direction of the field. It's one thing to say if the direction of the field is only in $\hat{r}$ then that is all fine and understandable (if I did the integral in cylindrical coordinates this would work wonderfully). But, if I had 3 components to the field direction would I still ignore the other two in favor of r?

The point is that $\hat r \cdot \hat \theta=0$. That's why you can ignore it. dA only has an $\hat r$. Physically the intuition is that the $\hat \theta$ component of the field isn't going into or out of the sphere, it's only going around. It doesn't matter for flux.

 

[bowlbase]

I get it, sort of. I understand the math but when I think of flux I think of everything, no matter direction, going through a surface. I want to count all of it, not just what is perpendicular to the surface.

I've finished the problem and it matches perfectly with the divergence of E and Gauss's theorem. Thanks a lot for the help guys!

 

[Dick]

I get it, sort of. I understand the math but when I think of flux I think of everything, no matter direction, going through a surface. I want to count all of it, not just what is perpendicular to the surface.

I've finished the problem and it matches perfectly with the divergence of E and Gauss's theorem. Thanks a lot for the help guys!

Great! You can count all of it, but the part of the field that's parallel to the surface is not going to contribute to the flux. Only the perpendicular part will. That's what the 'dot product' part says. Keep thinking about it.