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If I have two conducting, coaxial cylinders as shown:

enter image description here

The potential at the outer cylinder is equal to zero. And I apply a potential difference across both cylinders in the form $V_a - V_b = V_{ab}$, how can I find an expression of the potential everywhere, in terms of $V_{ab}$?

So this is what I have so far...

We place a $+Q$ on the inner cylinder and a -Q on the outer one. From Gauss' law, I know that $Q$ enclosed = $+Q$ so $\frac{Q}{\epsilon}$ holds.

I can set that equal to $2\pi rl E$ because $dA$ over all areas is just the surface area of the curved part of a cylinder. So $\frac{Q}{\epsilon} = 2rl\pi E$ and $E = \frac{Q}{\epsilon 2\pi rl}$ is the electric field between cylinders

Now, $E = \frac{dV}{dr}$ so the integral (from a to b) $dV$ is equal to $\Delta V_{ab} = -\frac{\lambda}{2\pi \epsilon}$

After doing all the integrals and bounds I got $\Delta V){ab} = -\frac{\lambda}{2\pi \epsilon}ln(\frac{b}{a})$

But this is only the potential between the two cylinders. I need the potential everywhere. So I still need to find the potential at the inside of the smaller cylinder and the potential on the outside of the bigger cylinder. Now I'm stuck!

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  • $\begingroup$ The link isn't working. Try posting it in the question directly. $\endgroup$
    – SuperCiocia
    Jul 26, 2015 at 21:58
  • $\begingroup$ The image should be working now $\endgroup$
    – user86788
    Jul 26, 2015 at 22:02
  • $\begingroup$ I assume the cylinders are infinitely long, or else this gets ugly. $\endgroup$
    – ragnar
    Jul 27, 2015 at 0:09
  • $\begingroup$ Also, please indicate specifically what you have tried and where you are getting stuck. Otherwise, you're just asking us to do your homework for you. $\endgroup$
    – ragnar
    Jul 27, 2015 at 0:20
  • $\begingroup$ Yes, I think we can make the assumption that they are infinitely long. Also, I edited the question to show where I got stuck. $\endgroup$
    – user86788
    Jul 27, 2015 at 2:15

2 Answers 2

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Consider starting with Laplace's equation in cylindrical, as this will give you the potential directly: $$ \frac{1}{r}\frac{d}{d r} \left(r\frac{d V}{d r}\right)=0 $$ since the space between the cylinder is charge-free. Moreover, you can use $d/dr$ rather than $\partial/\partial r$ since by symmetry $V=V(r)$ only.

It follows from this that $$ r\frac{dV}{dr}=C_1\qquad\Rightarrow\qquad V(r)=C_1\log(r)+C_2 $$ with $C_1$ and $C_2$ two integrating constants. You can find $C_1$ and $C_2$ using $V(r)$ at $a$ and $b$, and then use your expression for the potential difference to convert to an expression containing $\lambda$ etc.

I think you probably approached it right and found the correct potential difference between the cylinders but the next step is to integrate your $\vec E$-field from $a$ to $r$, where $r$ is a point inside the cylinder, so,as to get $V(r)-V(a)$. Since you are given a potential difference solving Laplace's equation is a little more direct.

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  • $\begingroup$ Please do not post complete or almost complete solutions to homework-like questions. Our policy on this can be found here which includes: “If someone posts an answer to a homework-type question that gives away a complete or near-complete solution, in most cases it will be temporarily deleted.” Please consider deleting this answer yourself. $\endgroup$
    – garyp
    Mar 18, 2017 at 12:10
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As you have figured out, $V_{ab} \equiv V_a - V_b = - \int^a_b \frac{\lambda}{2 \pi \epsilon_0r} = \frac{\lambda}{2 \pi \epsilon_0} ln(\frac{a}{b})$. But we don't know what $\lambda$ is yet! In fact, we must use this equation to derive its value in terms of $V_{ab}$, which is assumed known. Then, to get the potential at any radius $r$, simply do your integral of the field again, but this time let the limit be $r$ instead of $a$. Therefore, $V(r)=\int_b^r\frac{\lambda}{2 \pi \epsilon_0r}$ (no minus sign due to change of integration direction) $=\frac{\lambda}{2 \pi \epsilon_0} ln(\frac{r}{b})$, where $\lambda = \frac{2 \pi \epsilon_0 V_{ab}}{ln(\frac{a}{b})}$. Hope this helps!

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