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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?

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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.

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  • $\begingroup$ Thanks, for the answer , I understood the solution, however I fail to understand why my solution is incorrect, if we replace the conducting plane with a similar long straight wire parrallel to the original one, then the method above gives correct answer, in this case the field lines emerging from the original cylinder are not perepndicular to the surface of the other cylinder, I am confused $\endgroup$
    – green_32
    Apr 29, 2021 at 14:15
  • $\begingroup$ @green_32 The field you have considered is that of a straight wire, and such a wire generates a radial field, centered in the wire. At the plate's surface this field is not perpendicular to the plate, and thus it doesn't meet the boundary conditions for a conductor. Hence, your solution is incorrect. Said in another way, your choice of field is not a solution of the Laplace's equation with the boundary conditions given by that system of conductors. $\endgroup$ Apr 29, 2021 at 18:45
  • $\begingroup$ Instead, 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, and thus it meets automatically the required boundary conditions. $\endgroup$ Apr 29, 2021 at 18:45

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