Potential in spherical shells

[per persson]

Homework Statement

Two concentric spherical metal shells have radius a and 2a. The inner shell has potential V_o and the outer shell is grounded. In r<a, a<r<2a and r>2a there is vacuum. Determine the potential everywhere.

Relevant Equations

$V(r)=\int^{r}_{O}E\cdot dl$

In r<a the potential is V_o.


I don't understand why in a<r<2a, V(r)=V_o-\int^r_a E*dl. I would write V(r)=\int^r_a E*dl+\int^2a_\infty E*dl

I dont know how to write math symbols here but I wrote question here https://physics.stackexchange.com/questions/827681/upper-and-lower-bounds-in-integral-for-potential which is clearer.

 

[kuruman]

I dont know how to write math symbols here

To learn how to do it here, please click "LaTeX Guide" lower left above "Attach files".

The general expression for the potential function is $$V(r)-V(\text{ref})=-\int_{\text{ref}}^r \mathbf E \cdot ~d\mathbf r$$ where the potential is zero at $r=\text{ref}.$ Here the potential is zero at $r=2a$.
You can adapt this expression to (a) region II, $2a\leq r \leq \infty$ and (b) region I, $0\leq r \leq a$.

It would be easier not to involve integrals and invoke the Uniqueness theorem. Just write $$
V(r) =
\begin{cases}
A_I+\dfrac{B_I}{r}, & 0\leq r \leq a \\
A_{II}+\dfrac{B_{II}}{r}, & 2a\leq r \leq \infty \\
\end{cases}$$ and impose the boundary conditions at $r=a$, and $r=2a$ to find the constants.

 

[jbriggs444]

I don't understand why in a<r<2a, $V(r)=V_o-\int^r_a E*dl$ I would write $V(r)=\int^r_a E*dl+\int^{2a}_\infty E*dl$

I've put ## $\LaTeX$ tags on your formula above and supplied braces around the $2a$ superscript.

As written, you are integrating inward from infinity toward the outer shell and outward from the inner shell to the selected radius. That makes little sense to me. The formula would yield the difference in potential differences across two disconnected intervals.

By the shell theorem, we can quickly see that if the potential at $r=2a$ is zero then the potential everywhere outside with $r > 2a$ is also zero. It follows that $E$ is zero in the exterior of the outer shell and $\int^{2a}_\infty E * dl = 0$. It also follows that the net charges on the two shells are equal and opposite.

If you want the potential at $r$, you could use the reference potential of $0$ at $2a$ and integrate inward. Or you could use the reference potential of $V_o$ at $a$ and integrate outward.

Both require knowing $E$. So I would choose to do neither. [E is calculable from the givens here, but we have no need to do the calculation as long as we already understand that spherical shells have $\frac{1}{r}$ potentials]

Consider the potential in the absence of the outer shell. It is $K$ at $r=a$ and $\frac{K}{2}$ at $r=2a$. The potential difference between the two shells is $\frac{K}{2}$.

Consider the potential difference in the presence of the outer shell. The potentials in the region from $r=a$ to $r=2a$ will all be offset by a fixed constant from the above. The potential differences within the region will remain unchanged.

In particular, the potential difference from $r=a$ to $r=2a$ will remain unchanged at $\frac{K}{2}$. Since we know the potential at the outer shell is zero, this means that the potential at the inner shell will be $\frac{K}{2} = V_o$. It follows that $K = 2V_o$.

The potential function from just the inner shell will be $P = \frac{Ka}{r} = \frac{2V_oa}{r}$.

If we look at the boundary conditions (the known potentials at $r=a$ or $r=2a$) we can compute that the offset for the presence of the outer shell will be $\frac{K}{2}$.

The potential function from both shells combined will be $P=\frac{Ka}{r} - \frac{K}{2} = \frac{2V_oa}{r} - V_o$. For $a \le r \le 2a$.

By no coincidence this fits our boundary conditions of $V_o$ at $r=a$ and $0$ at $r=2a$.