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

• Queequeg
In summary, the potential difference between the inner and outer wire is calculated to be ##V=\frac{\lambda}{2\pi ε_0}\ln{\frac{R_1}{R_2}}## The capacitance of the wire is found to be ##C=\frac{Q}{V}## The total charge stored on the cable is found to be ##Q=\lambda L## The amount of stored energy in the cable is found to be ##κU## If the insulator changes from a vacuum to polystyrene, the new stored energy, capacitance, and charge density are found to be ##κC, 2.6, and \lambda respectively.
Queequeg

## 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}##?

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.

1 person
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

## What is the purpose of calculating potential, capacitance, charge, and energy in concentric wire homework?

The purpose of calculating these quantities is to understand the behavior of electric fields and how they affect charge distribution in a system. This knowledge can be applied to various real-world scenarios, such as designing electrical circuits or analyzing the performance of electronic devices.

## How is potential calculated in a concentric wire system?

Potential is calculated using the formula V = kQ/r, where V is the potential, k is the Coulomb constant, Q is the charge, and r is the distance from the charge. In the case of concentric wires, the potential can also be calculated by adding the potentials from each individual wire, taking into account their respective radii and charges.

## What is capacitance and how is it related to concentric wires?

Capacitance is the ability of a system to store charge. In concentric wire systems, capacitance is directly proportional to the ratio of the radii of the two wires. This means that as the radii of the wires increase, the capacitance also increases.

## How is charge distributed in a concentric wire system?

In a concentric wire system, the charge is distributed along the surface of the inner and outer wires, with the majority of the charge concentrated on the outer wire. This is due to the electric field being stronger on the outer wire, causing more charge to accumulate there.

## How is energy related to concentric wires and how can it be calculated?

Energy is related to concentric wires through the potential energy stored in the system. This energy can be calculated using the formula U = (1/2)CV^2, where U is the energy, C is the capacitance, and V is the potential difference between the two wires. This energy can also be calculated by multiplying the charge on each wire by the potential difference between them.

• Introductory Physics Homework Help
Replies
23
Views
533
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
220
• Introductory Physics Homework Help
Replies
4
Views
923
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
64
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
859
• Introductory Physics Homework Help
Replies
7
Views
314
• Introductory Physics Homework Help
Replies
10
Views
2K