Question

Two coaxial metal cylindrical tubes, of radius a and b > a, represent a cylindrical cable. The inner conductor at a carries a linear charge λ and the outer conductor at b is grounded. Assume an arbitrary length L. a) Determine the electric field in the region between the two conductors. b) Determine the energy stored in the system. Recall W = ε0/2 ∫ E^2 dτ, where the integral is over all space. c) This system also represents a cylindrical capacitor. You may recall the energy stored in a capacitor is given by W = Q^2 / (2C), where Q is the total charge and C is the capacitance. Determine the capacitance per unit length of the cable. 

          Two coaxial metal cylindrical tubes, of radius a and b > a, represent a cylindrical cable. The inner conductor at a carries a linear charge λ and the outer conductor at b is grounded. Assume an arbitrary length L.

a) Determine the electric field in the region between the two conductors.
b) Determine the energy stored in the system. Recall W = ε0/2 ∫ E^2 dτ, where the integral is over all space.
c) This system also represents a cylindrical capacitor. You may recall the energy stored in a capacitor is given by W = Q^2 / (2C), where Q is the total charge and C is the capacitance. Determine the capacitance per unit length of the cable.
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Two coaxial metal cylindrical tubes, of radius a and b > a, represent a cylindrical cable. The inner conductor at a carries a linear charge λ and the outer conductor at b is grounded. Assume an arbitrary length L. a) Determine the electric field in the region between the two conductors. b) Determine the energy stored in the system. Recall W = ε0/2 ∫ E^2 dτ, where the integral is over all space. c) This system also represents a cylindrical capacitor. You may recall the energy stored in a capacitor is given by W = Q^2 / (2C), where Q is the total charge and C is the capacitance. Determine the capacitance per unit length of the cable.
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Transcript

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00:02 So if we use u equals 1 half epsilon knot e squared, which is equal to one half epsilon not times 2 pi epsilon not of r squared, and that's equal to 8 pi squared, epsilon not r squared.
00:28 So then for part b we have u equals the integral of udv, which is equal to 2 pi l times the integral of ur, dr, which is l over 4 pi epsilon not, times the integral of rb r not times dr over r and u over l equals over 4 pi epsilon times the natural log of rb divided by ra...
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