# 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?

Finally, by assuming $$b\gg a$$, you are actually assuming an infinitely thin wire, for which the surface is orthogonal to any field direction.