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Chapter 26: Problem 11

A 50.0 -m length of coaxial cable has an inner conductor that has a diameter of \(2.58 \mathrm{mm}\) and carries a charge of \(8.10 \mu \mathrm{C} .\) The surrounding conductor has an inner diameter of \(7.27 \mathrm{mm}\) and a charge of \(-8.10 \mu \mathrm{C} .\) (a) What is the capacitance of this cable? (b) What is the potential difference between the two conductors? Assume the region between the conductors is air.

Short Answer

Expert verified
Solving the problems using the respective formulas and the given values gives us the capacitance and the potential difference between the conductors.

Step by step solution

01

Find the Capacitance

Given the diameters of both the inner (\(2.58 \, mm\)) and the outer conductor (\(7.27 \, mm\)), they are respectively converted to radii by dividing them by 2, and to meters by multiplying them by \(10^{-3}\). The length of the cable is given as \(50.0 \, m\). The formula \(C = \frac{2\pi\epsilon_0l}{ln(b/a)}\) is used for calculation where \(\epsilon_0 \approx 8.85 x 10^{-12} \, m^{-3}kg^{-1}s^4A^2\) is the permittivity of free space. This will give the value of capacitance in Farads (F).
02

Calculate the Potential Difference

The given charge on the inner radial conductor of the coax cable is \(8.10 \, \mu C = 8.10 x 10^{-6} \, C\). This is substituted into the formula \(V = \frac{q}{C}\), which yields the potential difference between the conductors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance Calculation
To understand how the capacitance is calculated in a coaxial cable, let's break it down. A coaxial cable's structure consists of two conductors, an inner and an outer, separated by a medium (usually air or an insulating material). Capacitance itself refers to the cable's ability to store electrical energy between these conductors. Building on the formula for the capacitance of a coaxial cable, we use \[ C = \frac{2\pi\epsilon_0l}{\ln(b/a)} \]This formula solves for the capacitance per unit length of the cable. Here's a bit more about its components:
  • \(\epsilon_0\) is the permittivity of free space, which is a constant.
  • \(l\) is the length of the cable.
  • \(b\) and \(a\) are the radii of the outer and inner conductors, respectively. Convert diameters to radii by dividing by two.
By applying these measurements to the formula, you can find the capacitance of the cable, allowing it to effectively carry electrical charge over its designated length.
Potential Difference
The potential difference, or voltage, between the two conductors of a coaxial cable, is a measure of the energy required to move a charge from one conductor to another. This concept is crucial for understanding how energy is distributed along the cable. The formula \[ V = \frac{q}{C} \]defines voltage, where:
  • \(V\) is the potential difference in volts.
  • \(q\) is the charge on the inner conductor.
  • \(C\) is the capacitance calculated earlier.
By substituting the known values of charge and capacitance, you get the potential difference in volts. This value is crucial for assessing the cable's performance, especially in telecommunications, where maintaining a consistent voltage is key to signal integrity.
Permittivity of Free Space
The permittivity of free space, denoted as \(\epsilon_0\), is a fundamental physical constant. It plays a pivotal role in defining how an electric field affects and is affected by a vacuum (in this case, the space between conductors in a coaxial cable). This value is approximately \(8.85 \times 10^{-12} \, F/m\). It essentially quantifies the ability of the vacuum to permit electric field lines, affecting how capacitors store charge.Understanding \(\epsilon_0\) is essential when calculating the capacitance of systems like coaxial cables. The permittivity of free space helps set the baseline for the electrical properties of the insulating material within the cable, which could be air, as in this exercise. The closer this material's permittivity is to free space, the closer the results will reflect theoretical calculations.

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Most popular questions from this chapter

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where \(\epsilon=\kappa \epsilon_{0}\) is the permittivity of the material. (a) \(\mathrm{A}\) sheet with charge \(Q\) uniformly distributed over its area \(A\) is surrounded by a dielectric. Show that the sheet creates a uniform clectric ficld at nearby points, with magnitude \(E=Q / 2 A \epsilon\). (b) Two large sheets of area \(A\), carrying opposite charges of equal magnitude \(Q,\) are a small distance \(d\) apart. Show that they create uniform electric field in the space between them, with magnitude \(E=Q / A \epsilon\) (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential Qd/Ae. (d) Show that the capacitance of the pair of plates is \(A \epsilon / d=\kappa A \epsilon_{0} / d\).

A \(10.0-\mu \mathrm{F}\) capacitor has plates with vacuum between them. Each plate carries a charge of magnitude \(1000 \mu \mathrm{C}\). A particle with charge \(-3.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-16} \mathrm{kg}\) is fired from the positive plate toward the negative plate with an initial speed of \(2.00 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Does it reach the negative plate? If so, find its impact speed. If not, what fraction of the way across the capacitor does it travel?

A wafer of titanium dioxide \((\kappa=173)\) of area \(1.00 \mathrm{cm}^{2}\) has a thickness of \(0.100 \mathrm{mm}\). Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a \(12.0-\mathrm{V}\) battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part (b), what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?

A 1 -megabit computer memory chip contains many \(60.0-\mathrm{fF}\) capacitors. Fach capacitor has a plate area of \(21.0 \times 10^{-12} \mathrm{m}^{2} .\) Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The order of magnitude of the diameter of an atom is \(10^{-10} \mathrm{m}=0.1 \mathrm{nm} .\) Express the plate separation in nanometers.

Show that the energy associated with a conducting sphere of radius \(R\) and charge \(Q\) surrounded by a vacuum is \(U=k_{e} Q^{2} / 2 R\).
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