# What does it mean to connect something to a potential V?

1. May 30, 2015

### carllacan

Hi.

I know how to work with potentials and how to solve problems involving them, but what exactly does it mean that we connect something (say, a spherical shell) to a potential V0? What is "actually" happening?

My guess is charges are "pumped" from some source to the material until an electric field in the radial direction is created such that the potential difference between the shell and infinity is V0. But if I were asked to calculate the expression for that electric field I would be at a loss, since I cant use E = -∇V.

2. May 31, 2015

### Simon Bridge

Physically it means that we arrange for the potential in the positions occupied by the shell to have the stated potential ... this could be done by hooking it to the positive terminal of a power supply or by putting it inside a carefully designed capacitor or some other means. The actual methods can be arbitrarily complicated. We put it like that because we don't want you to get distracted by the details of the means used to achieve that.
Why not?

3. May 31, 2015

### carllacan

Sorry, I actually meant to say I couldn't get the displacement field D.

The only way to obtain it is to know the polarization of the dielectric or the free charges inside, and in this case I know neither, so I don't know how to go from the potential to D, unless I somehow go first from the potential to the free charges.

4. May 31, 2015

### Simon Bridge

Note: The polarization comes in response to an electric field - what is the electric field inside a uniformly charged spherical shell?

You can calculate the potential everywhere outside the shell - from there you get the free charge distribution on the shell.
It may help to consider Poisson's equation.

5. Jun 1, 2015

### carllacan

Oh, that sounds really good. I had thought of using Poisson (or image charges) to find V outside the shell, but not to get the actual charge distribution.

Let me check if I have it right: I need to find V(r) outside the shell using Poisson (or image charges) and then equate it to the V'(r) that would be produced by a charge density σ in the shell. Once I have that I can get σ and then I can "forget" about the potential V0 and do the rest as if I had been given σ from the beginning. What troubles me is: can I really forget about V0? Have I, by means of getting σ, accounted for all the ways it interacts with the dielectric?

6. Jun 2, 2015

### Simon Bridge

You issue is more that you have not properly described the situation ... you started just by saying that a conducting shell is connected to a potential V0.
i.e. $V(R)-V(\infty) = V_0$ where R is the radius of the shell. You never said anything about what was inside the shell.

No need to solve the DE - it's been done for you many times: For a spherical shell charge Q and radius R, the potential is given by:
$V(r>R)=Q/4\pi\epsilon_0 r$

Given $V(R)=V_0$ and $r=R$, solve for Q:
$Q=4\pi\epsilon_0 RV_0$

Note: The electric field inside a closed conducting shell is zero - so there is no reason to find any polarization charges there.