Defining electrostatic potential field

1. Dec 28, 2016

oneofmany850

1. The problem statement, all variables and given/known data
In a device called a betatron, charged particles in a vacuum are accelerated by the electric
field E that necessarily accompanies a time-dependent magnetic field B(t).
Suppose that, in cylindrical coordinates, the magnetic field throughout the betatron at
time t can be approximated by
$$B = Kr^nt e_z$$
where K and n are positive constants. Assuming that the companion electric field takes the
form
$$E = E_φ(r) e_φ$$
express E in terms of r, K and n.
What mathematical feature of the electric field makes it impossible to define an
electrostatic potential field in this case? Explain your answer in two or three sentences.

2. Relevant equations

3. The attempt at a solution
$$E= -\frac{1}{2} Kr^{n+1} e_\phi$$
I'm not sure which feature of the field makes it impossible to define an electrostatic potential field. I'm thinking it might be that the E field is not conservative but I'm not sure why. Also the field is being generated by moving charges so I'm not sure how that can generate an electrostatic field. I wondered if the power of r had something to do with it.?

2. Dec 28, 2016

BvU

Hi,
Think you're doing fine. First part is answered. And you're thinking in the right direction. But
isn't the idea with betatrons. They get their field from some external current supplier. So stick to the conservativeness. In integral form it gets you a potential function. What's the derivative form for the comfortable choice of coordinates here ?

3. Dec 28, 2016

oneofmany850

The question asks for just a few sentences as explanation which is throwing me off thinking they don't want me to do an operation on the E field to determine if the field is conservative by E=-gradV or to show CurlE=0. I'm assuming it can be done by inspection. I have this written down so far:
The circular field lines mean that is has a non zero circulation and so it is not a conservative E field. Thus CurlE is not equal to zero and so the region is not simply connected therefore we cannot define an electrostatic potential field.
Is there a feature of the E field equation which I can use to justify it not being a simply connected region?

I should add that the field lines being circular is inferred from another part of the question which suggests this by asking if they point clockwise or anticlockwise from a particular viewpoint. I don't think this is the right way to go about this question though.

4. Dec 28, 2016

BvU

Check your Maxwell ! $\nabla \times \vec E$ can be nonzero ($\vec B$ varies in time !). But $\nabla\cdot\vec E$ should be zero. Hence my
for $\vec E = E_φ(r) e_φ$

5. Dec 28, 2016

oneofmany850

So $$divE = \frac{1}{r}\frac{d}{d\psi}-\frac{1]{2}Kr^{n+1} = 0$$
It equals zero as K, r and n are constants and there is no psi component to differentiate. Thus the divergence being zero implies that it is rotational and therefore is not conservative so we cant define an electrostatic potential field?

6. Dec 28, 2016

BvU

Well now! How did you determine this divergence ?
$$\nabla\cdot\vec E = \frac{1}{r}\frac{d}{d\psi}-\frac{1}{2}Kr^{n+1} = 0 \ \ ?????$$

7. Dec 28, 2016

oneofmany850

I think I'm dong the divergence in cylindrical coordinates incorrectly. I assumed because the E field is given for just the unit vector e_\phi then E_r and E_z equal zero. So just doing the partial differentiation on the phi component of E gives zero as there is no phi to differentiate?? Or am I supposed to differentiate r here?

8. Dec 28, 2016

BvU

Have to come back on that. How do you go from the given $B_z$ to $E_\phi$ ?

And what you meant to typeset in $\LaTeX$ was of course
$$\frac{1}{r}\frac{\partial }{\partial\phi}\left (-\frac{1}{2}Kr^{n+1} \right)$$ wasn't it ? And it's zero alright.

Only I got $E_\phi = - \displaystyle {Kr^{n+1}\over n+2}$ -- small detail. Divergence ${1\over r} {\partial E_\phi\over \partial \phi}$ is zero.​

And we can't write $\vec E$ as a gradient of a potential because the curl of a gradient is zero and the curl of this $\vec E$ is not. The non-conservativeness you started out with.

9. Dec 28, 2016

oneofmany850

the line integral of E equals the negative derivative of the surface integral of B. So

$$E_\phi 2\pi r = -\frac{d}{dt}[Kr^n t \pi r^2] = -K\pi r^{n+2}$$
so $$E_\phi = -\frac{1}{2}K r^{n+1}$$ since we took the 2pi r over.

Yes the phi component of the divergence was meant to be as you typed. But I cant find any examples of taking the divergence of a vector field in cylindrical coordinates so I'm not sure why it equals zero. But the last statement you made is helpful and explains the answer.

10. Dec 28, 2016

oneofmany850

oh I see lol I forgot to do the integral. So yes your answer is correct. But why does the divergence of this equal zero and how do we know the curl of E is zero also?

11. Dec 28, 2016

BvU

Re
When I google "divergence in polar", I end up here

And there I found the curl for the Maxwell eq $\nabla\times \vec E = -{\partial \vec B\over \partial t}$ (so the curl of $\vec E$ is not zero!), as well as ${1\over r}{\partial E_\phi\over \partial \phi}$ for $\nabla\cdot\vec E$, which is zero.

12. Dec 29, 2016

oneofmany850

okay, I think I have finally wrapped my head around it now haha. Thanks for all your help. I see that it is because this magnetic field is time dependent which means that the curl of E is not zero which means that we cant write the E field as a gradient since the curl of a gradient is zero. This leads us to the conclusion that this E field is non conservative because the electric field is irrotational and it produces circular field lines and thus defining an electrostatic potential field in this case is impossible.