Mutual capacitance of a long straight wire and an infinite conducting plate Q. A long straight wire is located parallel to an infinite conducting plate. The wire cross-sectional radius is equal to $a$, the distance between the axis of the wire and the plane equals $b$. Find the mutual capacitance of this system per unit length of the wire under the condition $b>>a$.
Consider a point at a distance $r$ from the axis of the cylinder, since we know that electric field $E(r)$ due to infinite line of charge is
$$E(r) = \frac{\lambda}{{2{\pi}{\epsilon}r}}.$$
So the potenital difference across the surfaces at a distance $a$ and $b$ respectively from the axis of the cylinder  comes to be
$$ V= \frac{\lambda \ln (b/a)}{2\pi\epsilon}.$$
So the capcitance is obtained by $\frac{Q}{V}$ . However, this method seems to give incorrect answer. I am not able to understand what I have applied incorrectly.
Note that I have not included the field to induced charges on the plate since the field due to the postive charge developed at the back and the negative charges on the front side of the plates cancel each other. Is this statement correct?
 A: The field right above a conducting surface should be perpendicular to the surface. Your choice of field does not meet this condition. That is, your choice of field is not a solution of the Laplace's equation with the boundary conditions set by the given system of conductors
To find the right field you can use the method of image charges, that is, you can remove the conducting plate if you imagine to have another long straight wire with opposite charge symmetrically placed with respect to the conducting plate. This ensures that the total field obtained in this way meets the boundary conditions at the conducting plate, because the superposition of the field generated by the original wire with that of its mirror image cancels the component of the field parallel to the plate, at the plate's surface.
The field with the plate will be equal to the one found in the above way, by means of the image charge, thanks to the uniqueness theorem.
Finally, by assuming $b\gg a$, you are actually assuming an infinitely thin wire, for which the surface is orthogonal to any field direction.
