Two long conducting wires of length L and radius a lie parallel a distance s apa

Question

Two long conducting wires of length L and radius a lie parallel a distance s apart (Fig. P2.8). The upper wire carries charge Q and the lower charge -Q. Since L >> s, we may assume the wires are effectively infinitely long for purpose of finding the electric fields and potentials. a) Find the electric field E in the plane in between the wires. b) Find the potential ! in the plane between the wires. Find V, the potential different between the wires. c) Find the capacitance C of the two wire system. d) Find the total electrical energy stored in the system.

Explanation / Answer

In order to find the capacitance per unit length between the two cylinders, we need to find the potentialdifference between them assuming that one carries a charge Q and the other Q. Since thecylinders supposed to be “long," we will say instead that each has a charge per unit length of , withQ=l, where l is the total length of the cylinder. Since we want capacitance per unit length in theend, this will be convenient.

Let us choose a coordinate system which has its origin on the center of the first cylinder of radius a1,which means the center of the second cylinder of radius a2 is located at r=d. Take the +ˆr directionto be along a line connecting the center of the two conductors toward the conductor of radius r2

Because the electric field obeys superposition, the total field at any point P is just the sum of the fieldsdue to each conductor separately. From Gauss’ law, we know that the field of each charged conductoris the same as that of a charged rod of length l and charge per unit length . Taking a point P alongthe axis connecting the center of the two conductors, we can find the total field readily:

E = /(2 r)

E total = /( r) { (1/r) +(1/(d-r)} r^

The potential difference between the two conductors can be found by integrating ~Etot · d~l over a pathconnecting the surface of the two conductors. Since ~E is conservative, we can take any path we like,and the most natural choice is to take a straight line path along ˆr along a line connecting the centersof the two conductors (horizontal dashed line above), viz., ˆr dr. Thus

V = Integral(a1 to d-a2) E tot dl = Q/(2 ) ln { (d-a1)(d-a2) /a1 a2}

For the last line, we noted that = Q/l. Now using the definition of capacitance, Q = CDV , orC=Q/DV , we have for the total capacitance

C= 2 l /{ln (a1a2/(d-a1)(d-a2)}

or, as asked, the capacitance per unit length C/l:

C/l = l /{ln (a1a2) /d}

given a1=a2=a hence , C/l = l /{ln a/d }

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