# How to create a uniformly charged sphere?

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## Summary:

Classical experimental electrostatics question: How to create a uniformly charged sphere and a uniformly charged spherical shell of finite thickness in the laboratory?
Can we create at least any one of the following in laboratory? How?

1. A uniformly charged spherical shell of finite thickness

2. A uniformly charged sphere

3. A radially symmetrically charged spherical shell of finite thickness

4. A radially symmetrically charged sphere

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vanhees71
Gold Member
2019 Award
The only thing I can imagine is to charge uniformly a spherical shell by just putting some charge on an isolated conducting sphere.

Will putting some charge on an isolated conducting sphere will make a spherical shell of finite thickness uniformly charged? Should the isolated conducting sphere needs to be inside the spherical shell of finite thickness?

vanhees71
Gold Member
2019 Award
The charge put on the sphere will uniformly spread on its surface (from a macroscopic point of view only of course; microscopically there's not smooth sphere to begin with).

The charge put on the sphere will uniformly spread on its surface (from a macroscopic point of view only of course; microscopically there's not smooth sphere to begin with).
Macroscopically, is there a way, at least, to make a spherical shell of finite thickness radially symmetrically charged?

jbriggs444
Homework Helper
2019 Award
Macroscopically, is there a way, at least, to make a spherical shell of finite thickness radially symmetrically charged?
An uncharged spherical shell of finite thickness has, trivially, a radially symmetric charge distribution.

A charged conducting spherical shell of finite thickness has a radially symmetric charge distribution with the unbalanced charge on a shell of infinitesimal thickness.

Your desire is for the charge density to be finite and non-zero throughout the volume of the spherical shell?

An uncharged spherical shell of finite thickness has, trivially, a radially symmetric charge distribution.
Can you explain why?

jbriggs444
Homework Helper
2019 Award
Can you explain why?
Well, what do you mean by a "radially symmetric charge distribution"?

I understand the term to be a synonym for "spherically symmetric" -- i.e. that the charge density at any point on a spherical shell of radius r is the same as at any other point on that shell.

oliverkahn
Well, what do you mean by a "radially symmetric charge distribution"?

I understand the term to be a synonym for "spherically symmetric" -- i.e. that the charge density at any point on a spherical shell of radius r is the same as at any other point on that shell.
Sorry, my connection was down.
Yes, I mean charge density at radius ##r## is constant

jbriggs444
Homework Helper
2019 Award
Sorry, my connection was down.
Yes, I mean charge density at radius ##r## is constant
A zero charge density everywhere is the same everywhere.

A zero charge density everywhere is the same everywhere.
No... I am talking about all radially symmetric charge distributions except that.

Your desire is for the charge density to be finite and non-zero throughout the volume of the spherical shell?
It cannot be zero for every ##r##.

jbriggs444
Homework Helper
2019 Award
No... I am talking about all charge distributions except that.
Then a charged conducting sphere does the job, right?

1) Why does a charged conducting sphere has radially symmetric charge distribution?

2) Or do you mean by placing a charged conducting sphere inside a spherical shell of finite thickness, the shell acquires radially symmetric charge distribution? Why?

vanhees71
Gold Member
2019 Award
The solution of Poisson's equation with appropriate boundary conditions is unique.

If you want spherical symmetry both the potential and the charge density must be functions of ##r## (in spherical coordinates only).

Thus 1) implies you have the solution
$$\phi(r)=\begin{cases} \frac{q}{4 \pi a} & \text{for} \quad r<a,\\ \frac{q}{4 \pi r} & \text{for} \quad r \geq a. \end{cases}$$
The charge distribution is
$$\rho=-\Delta \phi(r)=-\frac{1}{r} \partial_r^2 (r \phi) =-\frac{q}{4 \pi a r} \partial_r \Theta(a-r)=\frac{q}{4 \pi a^2} \delta(r-a).$$
This is a surface-charge distribution, ##\sigma=q/(4 \pi a^2)=\text{const}## along the sphere.

2) is somewhat more complicated. Let again ##a## be the radius of the inner conducting sphere and ##b## and ##c## the inner and outer radius of the shell of finite thickness, ##a<b<c##. For ##r<b## you have of course the same solution as above. Then you have
$$\phi(r)=\frac{q}{4 \pi} \begin{cases} 1/a & \text{for} \quad r < a, \\ 1/r & \text{for} \quad a \leq r<b, \\ 1/b & \text{for} \quad b \leq r<c, \\ 1/r & \text{for} \quad c \leq r. \end{cases}$$
You can calculate and interpret the various surface chrages on the surfaces of the conductors yourself!

oliverkahn
To summarize, I believe no one has suggested a way to do 1, 2, or 3. The only solution suggested for 4 is a charged conducting sphere where the charge will be uniformly distributed on the outer surface. The charge shell is effectively infinite thin, so this isn’t anything like what you asked for in 1-3, but you have to admit it is radially symmetric, and so technically a valid example that fits your #4 request (even if that isn’t what you meant to ask for)

In short, we haven’t helped much. I can’t think of a way to get a static charge distributed over a finite volume.

However, you do get quasi-static distributions of charge over volumes in dynamic situations, i.e. when current is flowing. A sharp pointed cathode inside a spherical anode will have a radially symmetric charge density. (1/r^2). A plate cathode and a plate anode will have a fairly uniform charge distribution between them (but not spherical, obviously). You could certainly make a cylinder of charge that way. A fluorescent light is a good example. Similar things can be done inside resistive or semiconducting materials. If you put point contacts on opposite sides of a spherical resister the charge density wouldn’t be a sphere and it wouldn’t be uniform, but it would certainly have shape. You might be able to tailor some interesting charge density shape by that kind of approach.

The closest I can think of to a uniformly charged ball is a plasma ball. These are the charged particle equivalent of smoke rings. The flow is toroidal, and the charge density isn’t uniform, but these can come close to being spherical. These occur naturally but they are created in the lab mostly for injecting plasma into fusion reactors. See Marshall injector or theta-pinch.

jbriggs444