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Chapter 22: Problem 37

22.37. The Coasial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius \(a\) and an outer coaxial cylinder with inner radius \(b\) and outer radius \(c\) . The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length \(\lambda\) . Calculate the electric field (a) at any point between the cylinders a distance \(r\) from the axis and \((b)\) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance \(r\) from the axis of the cable, from \(r=0\) to \(r=2 c\) (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

Short Answer

Expert verified
(a) Electric field is \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\) for \(a < r < b\). (b) Same formula for \(r > c\). (c) Graph shows decreasing hyperbola. (d) Charge per unit length is \(-\lambda\) on inner surface.

Step by step solution

01

Apply Gauss's Law

Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space, \(\varepsilon_0\). Mathematically, it is given by: \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \] For (a), a cylindrical Gaussian surface of radius \(r\) and length \(L\), where \( a < r < b \), is considered. The charge enclosed by this Gaussian surface is \( \lambda L \).
02

Calculate the Electric Field Between Cylinders (a < r < b)

By symmetry, the electric field \( \mathbf{E} \) is radial and has the same magnitude at every point on the Gaussian surface. The total electric flux \( \oint \mathbf{E} \cdot d\mathbf{A} \) through the Gaussian surface is \(E \cdot 2\pi r L\). Substituting in Gauss's Law gives: \[ E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0} \] Solving for \(E\) yields: \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \] Thus, the electric field between the cylinders is \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\).
03

Calculate the Electric Field Outside the Outer Cylinder (r > c)

For \(r > c\), consider a cylindrical Gaussian surface of radius \(r\) where \(r > c\). The total charge enclosed remains \(\lambda L\), since the outer cylinder has no net charge. The symmetry argument still applies, so \(E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0}\). Solving for \(E\) gives: \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \] Thus, the electric field outside the outer cylinder is \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\).
04

Graph the Electric Field Magnitude as a Function of r

The electric field \(E\) from the solution steps is a function of \(r\): \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\). This function is relevant for both regions, \(a < r < b\) and \(r > c\), and is depicted as a decreasing hyperbola. Graph \(E(r)\) starting from near zero, peaking immediately as it moves away from \(a\), decreasing across both regions, potentially with a drop-off at \(r = b\) due to shielding.
05

Determine Charge Distribution on Outer Cylinder

(d) Since the outer cylinder has no net charge, the charge per unit length induced on the inner surface of the outer cylinder is \(-\lambda\) due to charge separation caused by the inner cylinder's positive charge. On the outer surface of the outer cylinder, there is no additional charge, so it remains at \(0\).

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Key Concepts

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

Electric Field Calculation
To understand the electric field calculation in a coaxial cable, we use Gauss's Law. This principle reveals the relationship between electric flux and the charge it encompasses. In our scenario, we are evaluating a coaxial cable that consists of an inner charged cylinder and an outer uncharged cylinder. For the electric field between these cylinders, we choose a cylindrical Gaussian surface with radius \(r\) such that \(a < r < b\). Here, \(a\) and \(b\) represent the radii of the inner and outer cylinders respectively. Applying Gauss's Law, which states \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \), helps us understand how the charge \(\lambda \) over length \(L\) influences the electric field \(E\). This setup results in the equation \(E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0}\), ultimately giving \(E = \frac{\lambda}{2\pi \varepsilon_0 r}\).This equation indicates that the electric field is radial and inversely proportional to \(r\), highlighting its decrease as the distance increases.
Charge Distribution
Charge distribution along the coaxial cable drastically affects the electric field generated in and around the cable. The inner cylinder carries a positive charge per unit length, denoted as \(\lambda\). This singular positive charge induces a unique distribution across the cable.
  • On the inner surface of the outer cylinder: Charged particles on the inner surface of the outer cylinder are influenced by the electric field generated by the inner cylinder. The charge per unit length induced here is \(-\lambda\), essentially negating the effect of the inner cylinder's charge.
  • On the outer surface of the outer cylinder: Here, the charge does not change, remaining at \(0\) since the cable itself is neutral overall.
This separated charge distribution underlines how the charge on the inner cylinder is balanced by the induced charge on the inner surface of the outer cylinder, leaving the outer surface unaffected. This arrangement ensures the stable operation of the cable within different contexts.
Coaxial Cable
A coaxial cable is an essential component employed commonly in electrical and communication networks. It consists of an internal cylindrical conductor insulated from an external hollow cylindrical conductor.
  • Inner Conductor: Typically, the inner conductor is a solid wire that carries a signal or charge which often requires measurement or manipulation like in Gauss's Law application.
  • Outer Conductor: This segment acts both as a return path for the current and a shield to protect from external electromagnetic interference.
  • Insulating Layer: Situated between the two cylinders, it ensures that the inner and outer conductors remain electrically isolated while allowing physical stability.
Due to this unique setup, coaxial cables are revered for effectively managing electromagnetic fields around them, crucial for maintaining signal integrity in communication systems. The clear segregation within its structure ensures minimal signal loss, which is vital for efficient data transmission.

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Most popular questions from this chapter

22.48. Negative charge \(-Q\) is distributed uniformly over the sur- face of a thin spherical insulating shell with radius \(R\) . Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located (a) a distance \(r>R\) from the center of the shell (outside the shell) and (b) a distance \(r

22.31. A negative charge \(-Q\) is placed inside the cavity of a hol- low metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not?(c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the ciffects of the charge \(-Q ?\) In principle, could the same thing be done for gravity? Why or why not?

22.10. A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{m}\) . What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radins (a) \(0.500 \mathrm{m},(\mathrm{b}) 1.50 \mathrm{m},(\mathrm{c}) 2.50 \mathrm{m} ?\)

22.56. Can Electric Forces Alone Give Stable Rquilibrium? In Chapter \(21,\) several examples were given of calculating the force exerted on a point charge by other point charges in its sur- roundings. (a) Consider a positive point charge \(+q .\) Give an example of how you would place two other point charges of your choosing so that the net force on charge \(+q\) will be zero. \((b)\) If the net force on charge \(+q\) is zero, then that charge is in equilibrium. The equilibrium will be stable if, when the charge \(+q\) is displaced slightly in any direction from its position of equilibrium, the net force on the charge pushes it back toward the equilibrium position. For this to be the case, what must the direction of the electric field \(\overrightarrow{\boldsymbol{E}}\) be due to the other charges at points surrounding the equilibrium position of \(+q ?(c)\) Imagine that the charge \(+q\) is moved very far away, and imagine a small Gaussian surface centered on the position where \(+q\) was in equilibrium. By applying Gauss's law to this surface, show that it is impossible to satisfy the condition for stability described in part (b). In other words, a charge \(+q\) cannot be held in stable equilibrium by electrostatic forces alone. This result is known as Earnshaw's theorem. (d) Parts (a)-(c) referred to the equilibrium of a positive point charge \(+q\) . Prove that Earnshaw's theorem also applies to a negative point charge \(-q\) .

22.29. An infinitely long cylindrical conductor has radius \(R\) and uniform surface charge density \(\sigma\) . (a) In terms of \(\sigma\) and \(R\) , what is the charge per unit length \(\lambda\) for the cylinder? (b) In terms of \(\sigma\) , what is the magnitude of the electric field produced by the charged cylinder at a distance \(r>R\) from its axis? (c) Express the result of part (b) in terms of \(\lambda\) and show that the electric field outside the cylinder is the same as if all the charge were on the axis. Compare your result to the result for a line of charge in Example 22.6 (Section \(22.4 ) .\)
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