What is the potential differencebetween the cylinder axis and a point?

Question

Figure 3 shows a cross-section of the coaxialcylinders’ system consisting of a conducting cylinder
carrying linear charge density (lambda) and a conductingcylindrical shell carrying linear charge
density (lambda)1. (Linear charge densities here are the totalcharges of these bodies per unit length in
the axial direction perpendicular to the page.) The conductors areshown as filled shapes and
geometrical dimensions are indicated on the figure. What is thepotential difference V between
the cylinder axis O and point A outside the system at distance Rfrom the axis?

Here is the figure:
http://img44.imageshack.us/img44/2315/tempi.png
Figure 3 shows a cross-section of the coaxialcylinders’ system consisting of a conducting cylinder
carrying linear charge density (lambda) and a conductingcylindrical shell carrying linear charge
density (lambda)1. (Linear charge densities here are the totalcharges of these bodies per unit length in
the axial direction perpendicular to the page.) The conductors areshown as filled shapes and
geometrical dimensions are indicated on the figure. What is thepotential difference V between
the cylinder axis O and point A outside the system at distance Rfrom the axis?

Here is the figure:
http://img44.imageshack.us/img44/2315/tempi.png

Explanation / Answer

Given : 1) Use gauss's law to find the E-field between thecylinders ;b > r >a,
Use Gauss's law to find the E-field outside the cylinders; r >c
The field in the conductors is zero.
2) Now integrate Edr from the origin out to point A;

V(A) - V(0) = INTEGRAL(0>a)Edr + INTEGRAL(a>b)E(in)dr +INTEGRAL(b>c)Edr + INTEGRAL(c>A)E(out)dr

The first & third integrals are zero, since E = 0 in thoseregions.

The integrals of E(in) & E(out) will give your result. Theanswer will be logarithmic since you should have found E(in) &E(out) , from Gauss' Law, to be proportional to 1/ I hope it helps you

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