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Chapter 23: Problem 71

A transmission line consists of two parallel wires, of radius \(a\) and separation \(b,\) carrying uniform linear charge densities \(\pm \lambda\), respectively. With \(a \ll b\), their electric field is the superposition of the fields from two long straight lines of charge. Find the capacitance per unit length for this transmission line.

Short Answer

Expert verified
The capacitance of the transmission line per unit length is \(\pi\epsilon_0\).

Step by step solution

01

Compute the Electric Field

Recognize that the electric field \(E\) for a long line of uniform charge is given by \(E = \frac{\lambda}{2\pi\epsilon_0r}\), where \(r\) is the distance from the charge, \(\lambda\) is the linear charge density, and \(\epsilon_0\) is the permittivity of free space. Here, since there are two wires with charges \(\pm\lambda\) respectively, the electric field between the wires is the superposition of their individual electric fields. Since the charges are opposite and the field lines are in opposite directions, the resultant electric field in between the wires becomes \(E = \frac{\lambda}{\pi\epsilon_0 b}\).
02

Compute Potential Difference

The electric field \(E\) between the plates relates to the potential difference \(V\) between them by the equation \(E = \frac{V}{d}\), where \(d\) represents the separation between the charge distributions, which is \(b\) here. Solving for \(V\) gives \(V = E*b = \frac{\lambda b}{\pi\epsilon_0 b}\), the \(b\) cancels out and we get \(V= \frac{\lambda }{\pi\epsilon_0} \).
03

Compute the Capacitance

Capacitance \(C\) relates to charge \(Q\) and potential difference \(V\) by \(C = \frac{Q}{V}\). Here, \(Q = \lambda * l\), where \(l\) is the length of the transmission line per unit length. Hence substituting \(Q\) and \(V\) into the capacitance formula gives \(C = \frac{\lambda l}{\frac{\lambda }{\pi\epsilon_0}}\), with cancelling terms this simplifies to \(C= \pi\epsilon_0 \) per unit length.

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Most popular questions from this chapter

A parallel-plate capacitor has plates with area \(50 \mathrm{cm}^{2}\) separated by \(25 \mu \mathrm{m}\) of polyethylene. Find its (a) capacitance and (b) working voltage.

A sphere of radius \(R\) carries total charge \(Q\) distributed uniformly over its surface. Show that the energy stored in its electric field is \(U=k Q^{2} / 2 R\).

A camera requires \(5.0 \mathrm{J}\) of energy for a flash lasting \(1.0 \mathrm{ms}\). (a) What power does the flashtube use while it's flashing? (b) If the flashtube operates at \(200 \mathrm{V},\) what size capacitor is needed to supply the flash energy? (c) If the flashtube is fired once every \(10 \mathrm{s},\) what's its average power consumption?

A parallel-plate capacitor is connected to a battery that imposes a potential difference \(V\) between its plates. If a dielectric slab is inserted between the plates, what happens to (a) the potential difference, (b) the capacitor charge, and (c) the capacitance?

A car battery stores about \(4 \mathrm{ M J }\) of energy. If this energy were used to create a uniform \(30-\mathrm{kV} / \mathrm{m}\) electric field, what volume would it occupy?
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