haruspex said:
No, several errors.
You have written the expression for the field at some arbitrary point, (y, z).
First problem: the expression would be correct without the plate. Perhaps the charge distribution on the plate changes it?
Second problem: you integrate this wrt z,but leaving y as arbitrary. What path is that along?
Note that it would be simpler to run the integral in the other direction, avoiding the d-z.
Third problem: if you fix the second problem, you will get an integral that diverges at the wire end. You need to take into account the radius of the wire.
Fourth problem. You have z as a bound on the integral. What should it be?
Now, I correct my process!
Firstly, the electric field generated by the plane is E_{plane}=\frac{\sigma}{2\epsilon_{0}}\hat{\mathbf{z}}.
Then, by the definition of potential, I can get the potential V_{plane}=-\frac{\sigma}{2\epsilon_{0}}z
On the other hand, the electric field generated by the wire is E_{wire}=\frac{\lambda}{2\pi\epsilon_{0}r}\hat{\mathbf{r}}
Then, the potential is V_{wire}=-\frac{\lambda}{2\pi\epsilon_{0}}\ln{\frac{r}{a}}=-\frac{\lambda}{2\pi\epsilon_{0}}\ln{\frac{\sqrt{(z-d)^{2}+y^{2}}}{a}}
Thus, the total potential will be V_{total}=-\frac{\sigma}{2\epsilon_{0}}z-\frac{\lambda}{2\pi\epsilon_{0}}\ln{\frac{\sqrt{(z-d)^{2}+y^{2}}}{a}}
Now, we can use the relation \sigma=-\epsilon\frac{\partial V}{\partial z}|_{z=0} to get the surface charge density.
<br />
<br />
\sigma = \frac{\lambda}{2\pi} \frac{a}{\sqrt{(z-d)^{2}+y^{2}}} \frac{1}{2} \frac{2(z-d)}{\sqrt{(z-d)^{2}+y^{2}}} |_{z=0} + \frac{\sigma}{2}\\<br />
\frac{\sigma}{2} = \frac{\lambda}{2\pi} \frac{-ad}{d^{2}+y^{2}}\\<br />
\sigma = \frac{-\lambda a d}{\pi(d^{2}+y^{2})}<br />
<br />
Then, we can go back and get the total potential in the system
<br />
V =-\frac{\lambda}{2\pi\epsilon_{0}}\ln{\frac{\sqrt{(z-d)^{2}+y^{2}}}{a}} + \frac{\lambda adz}{2\pi\epsilon_{0}(d^{2}+y^{2})}<br />
and the capacitance per length of the wire is
<br />
C = \frac{\lambda}{V} = \frac{2\pi\epsilon_{0}(d^{2}+y^{2})}{adz} - 2\pi\epsilon_{0}\ln{\frac{a}{\sqrt{(z-d)^{2}+y^{2}}}}<br />
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