Concentric Wire Homework: Calculating Potential, Capacitance, Charge & Energy

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SUMMARY

The discussion focuses on calculating the potential difference, capacitance, charge, and energy stored in a concentric wire system with resistivity ##\rho## and charge density ##\lambda##. The potential difference between the inner wire (radius ##R_1##) and the outer conductor (radius ##R_2##) is derived using the formula ##V = \frac{\lambda}{2\pi ε_0}\ln{\frac{R_1}{R_2}}##. The capacitance is calculated as ##C = \frac{2πε_o L}{\ln{\frac{R_1}{R_2}}##, while the total charge stored is ##Q = CV##. The stored energy is determined using ##U = \frac{1}{2} CV^2##, and the impact of changing the insulator to polystyrene is analyzed, resulting in new capacitance and energy values based on the dielectric constant of 2.6.

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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}##?
 
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I feel you have it right.
You might benefit from a drawing of the system, to make sure the geometric is totally clear.
You need to explicit the cross-section A.
My understanding is that the current flows along the central conductor.
Its radius is R1, which determines its cross-section.
The outer conductor is probably grounded "everywhere" , V=0 all along its length, therefore it does not dissipate energy.
 
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Doesn't the inner wire dissipate energy though, so the fraction supplied to the cable is just all of the power dissipated in the inner wire, using ##I^2 R##?

Unless by "cable wire" that's the outer conductor in which case I see no power is dissipated.

Thanks
 

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