Answer to the Question 10/04
WIRE CAPACITORThe question was:
A capacitor consists of two very long parallel conducting circular wires almost touching each other (see the cross section in the picture). Find the electric field of the capacitor. Does it have a finite capacitance? Explain!
(8/05) The solution of the problem can be found in numerous books. J.I.I. de la Torre (e-mail nacho@usal.es). sent us (30/5/05) a detailed analysis of the solution that can be found in an editted form in the following PDF file. Below we briefly outline the solution.
The answer: If the minimal distance between the cylinders is d, then the capacitance per unit length is
2πε/cosh-1(1+2d/R+(d/2R)2),
where R is the radius of the cylinders, and ε is the permittivity of vacuum. When the separation d is much smaller than the radius of the cylinders R, the the capacitance is approximately πε(R/d)1/2. Not surprisingly, for d=0, the capacitance diverges.
The solution: In the plane perpendicular to the cylinders we are faced with a two-dimensional problem. This, and similar problems, are solved in many books considering two-dimensional electrostatics. In particular, it can be found in the Static and Dynamic Electricity by W.R. Smythe (Taylor&Frances, New York, 1989). (We thank K.T. McDonald for pointing out this reference.) Here, we outline the solution.
Electrostatic potential of two line charges 2πε and -2πε is given (in MKSA (SI) units) by the function
U=ln(r2/r1),
where r1 and r2 are the distances from the wires to a given point. Using, complex two dimensional plane with z=x+iy, with wires located at +ia, and -ia, we can write
U=Re ln((z+ia)/(z-ia)).
This relation can be re-written as
x2+(y-a coth U)2=a2 csch2 U.
From this relation we immediately see, that equipotential lines are circles centered at a coth U with radius a|csch U|. By choosing potentials of the cylinders U1 (negative) and U2 (positive) at the surfaces of the cylinders, i.e.
R=-a csch U1=a csch U2,
and the separation between the centers
2R+d=a (-coth U1+coth U2),
we find that
cosh(U2-U1)=1+2d/R+(d/2R)2.
By dividing the charge per unit length by potential difference, we find the capacitance quoted in the answer.
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