1. The problem statement, all variables and given/known data

A concentric wire of resistivity ##\rho##,length ##L## has an inner radius ##R_1## and outer radius ##R_2## and charge density ##\lambda##. A current ##I## flows down the inner wire while the outer conductor is grounded, ##V=0##.

a) Calculate the potential difference between the inner and outer wire.

b) Find the capacitance of the wire.

c) If the potential difference between the inner and outer conductor is ##V##, calculate the total charge stored on the cable.

d) Calculate the amount of stored energy in the cable.

e) If the insulator changes from a vacuum to polystyrene, what is the new stored energy, capacitance, and charge density?

f) What fraction of the power supplied to the cable is dissipated in the cable wire itself?

2. Relevant equations

##E=\frac{\lambda}{2\pi ε_0 r},
C = \frac{Q}{V},
Q = \lambda * L,
U = \frac{1}{2} CV^2,
P = IV = I^2 R##

3. The attempt at a solution

a) I integrate the electric field from the inner to outer cable to get ##V = \frac{\lambda}{2\pi ε_0}\ln{\frac{R_1}{R_2}}##

b)## Q = \lambda L## so ##C = \frac{Q}{V} = \frac{Q}{\frac{\lambda}{2\pi ε_0}\ln{\frac{R_1}{R_2}}} = \frac{2πε_o L}{\ln{\frac{R_1}{R_2}}}##

c) ##Q = CV = V * \frac{2πε_o L}{\ln{\frac{R_1}{R_2}}}##

d) Just ## U = \frac{1}{2} CV^2##

e) Polystyrene has a dielectric constant of 2.6, so the new capacitance is ##κC##, the energy is ##κU## and the charge density is the same, ##\lambda##

f) Not too sure, I’m guessing the total power provided is ##P = I\Delta V## and the power in the inner wire is ##I^2 R## so the fraction is just ##\frac{IR}{V}## where ##R=\frac{\rho L}{A}##?

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