The inner conductor of a coaxial cable has a radius of alpha = 0.8 mm, and the o

Question

The inner conductor of a coaxial cable has a radius of alpha = 0.8 mm, and the outer conductor's inside radius is b = 3 mm. The length of the cable is I = 1 m. What is the cable's capacitance if the space between the two conductors is empty? What is the cable's capacitance when the space between the conductors is filled with polyethylene, which has a dielectric constant of k = 2.3? Briefly explain what are the physical origins of the capacitance difference in (i) and (ii). Derive all the necessary expressions, starting from first principles (i.e. Gauss' Law, and the potential definition). The Coulomb constant is k_e = 8.987 middot 10^9 N m^2/C^2.

Explanation / Answer

consider a concetric cylinder of length L and radius r.

where L=1 m

and a<r<b

then charge enclosed by this surface=Q

using gauss law, if electric field is E,

then epsilon*E*2*pi*r*L=Q

==>E=Q/(2*pi*epsilon*r*L)

then potential difference=integration of -E*dr

magnitude of potential difference=integration of E*dr

=integration of Q*dr/(2*pi*epsilon*r*L)

=(Q/(2*pi*epsilon*L))*ln(r)

so magnitude potential difference between inner cylinder and outer cylinder

=(Q/(2*pi*epsilon*L))*ln(b/a)

as we know, for a capacitor, charge=capacitance*potential difference

==>capacitance=charge/potential difference

=Q/((Q/(2*pi*epsilon*L))*ln(b/a))

=2*pi*epsilon*L/(ln(b/a))

part a:

when the space is empty, epsilon=epsilon_0 =8.85*10^(-12) (electrical permitivity of free space)

then capacitance=2*pi*8.85*10^(-12)*1/(ln(3/0.8))

=4.207*10^(-11) F

part b:

when the conductor is filled with a dielectric,

electrical permitivity becomes dielectric constant times electrical permitivity of free space

so capacitance=2*pi*2.3*8.85*10^(-12)*1/(ln(3/0.8))=9.676*10^(-11) F

part c:

in part b, there is a dielectric present which increases the capacity to hold charge and in turn, the capacitance.

the multiplying factor is the dielectric constant of the dielectric.

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