- #1

Queequeg

- 25

- 0

## Homework Statement

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?

## Homework 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##

## 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}##?