What is the potential between 2 concentric spheres?

[Orodruin]

It never ceases to amaze me how you can say a relatively basic thing in so many ways.

If you have not yet solved it, since you seem to be at least somewhat familiar with integration and therefore should be with differentiation, I would suggest going back to ehild's post #16. You may or may not prefer the differential equation on the form
$$
r^2 U''(r) + 2r U'(r) = 0,
$$
with boundary conditions $U(r_1)=V_1$, $U(r_2)=V_2$. Once done it will be easier to discuss the solution and the interpretation of it.

 

[STEMucator]

It never ceases to amaze me how you can say a relatively basic thing in so many ways.

If you have not yet solved it, since you seem to be at least somewhat familiar with integration and therefore should be with differentiation, I would suggest going back to ehild's post #16. You may or may not prefer the differential equation on the form
$$
r^2 U''(r) + 2r U'(r) = 0,
$$
with boundary conditions $U(r_1)=V_1$, $U(r_2)=V_2$. Once done it will be easier to discuss the solution and the interpretation of it.

The S.I. unit system and the concepts of mathematics are a privilege to learn, and to apply.

That differential equation solution is quite interesting, but the simplicity of haruspex's post would probably be the elementary way of visualizing this.

 

[Orodruin]

The SI unit system has nothing to do with the matter. The problem as posed does not require a system of units to solve. Are you perhaps rather referring to the use of basic solutions such as that of a point charge or a spherical shell to solve the problem? My point is that several people, including ehild, haruspex, and myself, have been trying to find a way of getting that message through for the better part of two full pages of posts now. Solving the differential equation and use the solution to discuss the physics of it afterwards is my if-all-else-fails approach.

 

[ehild]

It never ceases to amaze me how you can say a relatively basic thing in so many ways.

If you have not yet solved it, since you seem to be at least somewhat familiar with integration and therefore should be with differentiation, I would suggest going back to ehild's post #16. You may or may not prefer the differential equation on the form
$$
r^2 U''(r) + 2r U'(r) = 0,
$$
with boundary conditions $U(r_1)=V_1$, $U(r_2)=V_2$. Once done it will be easier to discuss the solution and the interpretation of it.

The equation in Post #16 is easier to solve for r1<r<r2:

$\frac{1}{r^2}\frac {\partial }{\partial r}\left(r^2 \frac {\partial U}{\partial r}\right)=0$

Multiply by r²:

$\frac {\partial }{\partial r}\left(r^2 \frac {\partial U}{\partial r}\right)=0$

Integrate once, the right side is a constant: $r^2 \frac {\partial U}{\partial r}=A\rightarrow \frac {\partial U}{\partial r}=\frac{A}{r^2}$

Integrate again :$U=-\frac{A}{r}+B$.

Use the boundary conditions:

$V_1=-\frac{A}{r_1}+B$
$V_2=-\frac{A}{r_2}+B$

It is two equations for the constant A and B. Solve, substitute into the equation for U(r): $U=-\frac{A}{r}+B$. You can calculate the potential at any point between the shells.

ehild

 

[Orodruin]

The equation in Post #16 is easier to solve for r1<r<r2:

I guess this depends on your definition of "easy" and pre-knowledge. Recognizing it as an ODE of Euler type, it is just about insering $r^k$ and solving the resulting $k(k+1)=0$.

That said, neither method is really hard.

 

[Vibhor]

Is $\frac{V_2r_2^2-V_1r_1^2}{r_2^2-r_1^2}$ the answer ?

 

[ehild]

Is $\frac{V_2r_2^2-V_1r_1^2}{r_2^2-r_1^2}$ the answer ?

Not. How did you get it? I think you made some mistake by cancelling some terms.ehild

 

[Vibhor]

Sorry .

It should be $\frac{V_1r_1+V_2r_2}{r_1+r_2}$ . Right ?

 

[ehild]

Right :)

ehild

 

[Vibhor]

Thanks a lot !

 

[Orodruin]

Ok, since the cat is out of the box, let us reconnect to the physics and the more physical arguments way of doing the problem and see how it relates to the "shut-up-and-solve-the-ODE" approach.

If we can find a charge configuration in space which does not contain any charge between the shells and reproduces the correct potentials on the shells at $r_1$ and $r_2$, then the corresponding potential for points between the spheres will be the solution to the problem.
Since the problem is obviously spherically symmetric, we are only interested in spherically symmetric solutions with no charge between the spheres. Between the spheres, any spherically symmetric charge distribution at $r < r_1$ is going to produce a potential proportional to $1/r$ while any spherically symmetric charge distribution at $r>r_2$ is going to produce a constant potential according to the shell theorem (it is also possible to ignore the charge outside and just refer to potentials being defined up to a constant addition). Thus, the most general spherically symmetric solution between the spheres is given by
$$
V(r) = \frac Ar + B,
$$
with $A$ and $B$ being constants given by the charges.
Of course, this is exactly the same result that was obtained by solving the differential equation and the determination of $A$ and $B$ (and thus, the charges) is performed in the same way.

 

[ehild]

If you want an even more "Physical" solution, without differential equations, determine the electric field between the shells. Using symmetry again, we assume that E is radial and its magnitude is equal in any direction. Applying Gauss' Law at distance r from the centre,
$$\oint {EdA}=4πr^2E=Q_1/ε_0$$where Q₁ is the charge on the inner sphere (unknown yet) so $E=\frac{Q_1}{4\piε_0}\frac{1}{r^2}=kQ_1\frac{1}{r^2}$

The electric field is negative gradient of the potential, U. $\vec E(r)=-\nabla U(r)$.

The potential at a point between r1 and r2 is obtained by integrating the equation above between r1 and r:

$U(r)-U(r_1)= -kQ_1\int_{r1}^r{r^{-2} dr}= kQ_1(\frac{1}{r}-\frac{1}{r_1})$

U(r₁)=V₁ and U(r₂)=V₂, from these you get kQ₁:
$$kQ_1=\frac {V_2-V_1}{\frac{1}{r_2}-\frac{1}{r_1}}$$

and $U(r)=V(r_1)+ kQ_1(\frac{1}{r}-\frac{1}{r_1})$

 

[Red]

Thank you guy very much! Thank you haruspex, Orodruin, ehild and the rest for your help!

 

[shameem420]

Potential deferece between two coaxial cylindre