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!

  • $\begingroup$ The link isn't working. Try posting it in the question directly. $\endgroup$
    – SuperCiocia
    Commented Jul 26, 2015 at 21:58
  • $\begingroup$ The image should be working now $\endgroup$
    – user86788
    Commented Jul 26, 2015 at 22:02
  • $\begingroup$ I assume the cylinders are infinitely long, or else this gets ugly. $\endgroup$
    – ragnar
    Commented 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
    Commented 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
    Commented Jul 27, 2015 at 2:15

2 Answers 2


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.

  • $\begingroup$ 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
    Commented Mar 18, 2017 at 12:10

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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.