# Induced charge density problem

1. Oct 21, 2014

### skrat

1. The problem statement, all variables and given/known data
I actually have three problems that have one thing in common that I don't understand. I will try to shortly describe all of them:

1. We have two perpendicular planes. (See Attachment: sketch1.png) The idea is to describe the electric potential in space. Now you can believe me, that the electric potential is given as $0(r,\varphi )=\frac{2U_0}{\pi }\varphi$. The real trouble is to calculate the induced charge on the bottom plate.

2. In a homogeneous electric field $\vec{E_0}$ a ball (3D Sphere) is inserted. Again the idea is to calculate the electric potential anywhere in space and you can trust me that the electric potential is $U(r,\vartheta )=E_0\frac{a^3}{r^2}\cos \vartheta -E_0r\cos \vartheta$ if $a$ is the radius of the sphere. The problem is again to calculate the induced electric charge.

3. A charge above a metal plate. (See attachment Sketch2.png). Again believe me that the electric potential is $U(r,\vartheta )=\frac{e}{4\pi \varepsilon _0}(\frac{1}{\sqrt{r^2-2rd\cos \vartheta + d^2}}-\frac{1}{\sqrt{r^2+2rd\cos \vartheta + d^2}})$ if $d$ is the distance between the electron and the plate. The question is again: What is the induced charge density on the plate?

2. Relevant equations

$$\sigma _{IND} =-\varepsilon _0|\vec{ \nabla }U|$$

3. The attempt at a solution

Now I have the solutions, but the problem is that I don't understand them.

1. The $\vec{ \nabla }U$ has only one component in z direction.

$(\vec{ \nabla }U)_z=\frac 1 r \frac{\partial U}{\partial \varphi}$

2.

$|\vec{ \nabla }U|=-\varepsilon _0|\frac{\partial U}{\partial r}|$

3.

$|\vec{ \nabla }U|=-\varepsilon _0|\frac{\partial U}{\partial z}|=-\varepsilon _0|\frac 1 r \frac{\partial U}{\partial \cos \vartheta}|$

So... If somebody could please explain me how on earth did we get those last equations? How do I know in each case what exactly $|\vec \nabla U|$ is? :/

Last edited: Oct 21, 2014
2. Oct 26, 2014

### Greg Bernhardt

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Oct 27, 2014

### skrat

Not really ... No worries though, I will figure this out - simply because I have to.

4. Oct 28, 2014

### BvU

1. $U_0$ is probably the $v$ in the figure ? And $\phi$ is some angle, which I can imagine. With two (infinitely long in z direction ?) planes, I can't imagine a $z$ dependence, so I'm a little lost. Can you explain the coordinates ?

help me a little....

5. Oct 28, 2014

### skrat

That's $U$ in the figure but the figure is so small it looks like $v$.
Exactly. We wanted to get the potential in space between the plates - 1. quadrant if you want so. Both plates are infinitely long! Meaning we have to solve a laplace equation $\nabla ^2 U=0$ and we did it in polar coordinates where $r$ represents the distance from the origin and of course $\varphi$ is the angle - naturally $\varphi \in [0,\frac{\pi }{2}]$.

The solution is $U=\frac{2U_0}{\pi }\varphi$. Meaning the potenital is a function of $\varphi$ only. (Ok I changed my notation on the figure from $U$ to $U_0$. But I think you get the point.)

6. Oct 28, 2014

### BvU

I find it strange the dimensions of the plates don't come in. They don't look the same witdth to me, but perhaps it was a given that they are both infinitely wide as well: that would explain why U doesn't go to 0 for $r \rightarrow \infty$.

OK, so the Electric field is minus the gradient of U. The gradient is $${\partial U\over \partial r}\, \hat r + {1\over r} {\partial U\over \partial \phi}\, \hat \phi + {\partial U\over \partial z}\, \hat z = {1\over r} {\partial U\over \partial \phi}\, \hat \phi = {2U_0\over r \pi }\, \hat \phi$$ In other words: at the grounded plane the E field is pointing straight down. Not in the z direction but in the $-\phi$ direction. That's reasonable: if it weren't there would be a force component along the conductor that would move the charges until E is perpendicular again.

Now the step from E to $\sigma$ is made using Gauss' law, your relevant equation. You find an induced charge density that drops of as $1\over r$. Seems fine to me... total charge can't be calculated, but that's what you get for infinities.

Your case 3 is worked out in detail here

And, eh, not that I don't trust you, but the U you found in case 2 is of course only for $r>a$ ;)

And you will need an expression for the gradient in spherical coordinates.

Last edited: Oct 28, 2014
7. Oct 28, 2014

### skrat

Am... is it always that simple? I only have to write down the gradient in the right coordinates and that's it? There is no need for guessing or too much thinking?

About case 2: You are correct - that's what the problem was asking.

8. Oct 28, 2014

### BvU

It's not always that simple. And after all you've done a lot of work to find the expressions for U ...

9. Oct 28, 2014

### skrat

Perfect, thanks a lot BvU!