Fun with conductors

First thing you should use is the method of image. For example if you have a point charge and a plate at distance r potential in the space would be as same as if you have + and - point charge at double distance 2r .

Same way in this problem you have two cylindrical conductors at distance 2h Now by Gauss law we can find electric field created by one cylindar. Inside its 0 and outside its

${E(x)}=\frac{\lambda}{2\pi \epsilon_0 {x}}$ use that $E(x)=-\frac{dV}{dx}$

and now you have that the potential is

$V=-\frac{\lambda}{2\pi \epsilon_0} ln(x) +C$

Now we have to use method of image again. We will replace the cylindars with lines with a same charge per lenght and distance d from centar of cylindar. Also we now that potential is same at any point at surface of cylindar so we should find

$V=V(d, \phi)$ (at the surface of cylindar) and use $\frac{dV}{d\phi}=0$ to find d .

But you can make your life easier and say that $V(A)=V(B)$ (point A and B at picture).

$V(A)=-\frac{\lambda}{2\pi \epsilon_0} ln(2h-d-r)+\frac{\lambda}{2\pi \epsilon_0} ln(r-x)$ and $V(B)=-\frac{\lambda}{2\pi \epsilon_0} ln(2h-d+r)+-\frac{\lambda}{2\pi \epsilon_0} ln(r+x)$

...Now after little a bit of calculation you will find that

$d=h-\sqrt{h^2-r^2}$ (you have two solutions of quadratic equation but notice that d need to be smaller than r , you can proove it by using Gauss law) and finaly force per unit of lenght is

$\frac{dF}{dl}={E(x)}\lambda$ where $x=2(h-d)$

$\frac{dF}{dl}= \frac{\lambda}{4\pi\epsilon_0 \sqrt{h^2-r^2}}$

Sorry for bad English :)

\int{1}{2}