If I have two conducting, coaxial cylinders as shown:
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!
Best Answer
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.