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You may have seen a coaxial cable connected to a television set. As shown in Figure 21.69, a coaxial cable consists of a central copper wire of radiusr1 surrounded by a hollow copper tube (typically made of braided copper wire) of inner radiusr2 and outer radius r3. Normally the space between the central wire and the outer tube is filled with an insulator, but in this problem assume for simplicity that this space is filled with air. Assume that no current runs in the cable.

Suppose that a coaxial cable is straight and has a very long length L, and that the central wire carries a charge +Q uniformly distributed along the wire (so that the charge per unit length is +Q/L everywhere along the wire). Also suppose that the outer tube carries a charge -Q uniformly distributed along its length L. The cylindrical symmetry of the situation indicates that the electric field must point radically outward or radically inward. The electric field cannot have any component parallel to the cable. In this problem, draw mathematical Gaussian cylinders of length d (with d much less than the cable length L) and appropriate radius r, centered on the central wire.

(a). Use a mathematical Gaussian cylinder located inside the central wire(r<r1) and another Gaussian cylinder with a radius in the interior of the outer tube(r2<r<r3) to determine the exact amount and location of the charge on the inner and outer conductors. (Hint: What do you know about the electric field in the interior of the two conductors? What do you know about the flux on the ends of your Gaussian cylinders?)

Short Answer

Expert verified

The charge on the outer surface of the inner wire is +Q.

The charge on the inner surface of the outer wire is -Q.

Step by step solution

01

Given Information.

A coaxial cable consists of a central copper wire of radius r1surrounded by a hollow copper tube (typically made of braided copper wire) of inner radius r2and outer radius r3. Normally the space between the central wire and the outer tube is filled with an insulator, but in this problem assume for simplicity that this space is filled with air.

02

Definition/ Concept.

The electric field inside a conductor is zero. This can happen only when the charge given to a conductor resides only on its surface.

03

Determine the exact amount and location of the charge on the inner and outer conductors.

The figure will be:

A Gaussian cylinder constructed with a radius r such that r<r1, is a cylinder that resides that resides inside the cylindrical wire of radius r1. The electric field inside the wire is zero. This means that the charge +Q distributed along its length L cannot lie in its interior. It follows therefore that the entire charge lies on the outer surface of the conductor.

Therefore, the charge on the outer surface of the inner wire is +Q.

The figure will be:

A Gaussian cylinder of radius r and length d is constructed such that r2<r<r3. This is a cylinder that lies inside the outer cylindrical wire as shown above. The outer wire contains a charge -Q. But the field inside the outer wire is also zero. Therefore, since the field lines point radically, the charge -Q must reside on the inner surface of the outer cylinder.

Therefore, the charge on the inner surface of the outer wire is -Q.

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Most popular questions from this chapter

Based on its definition as electric flux per unit volume, what are the units of the divergence of electric field.

In chapter 15 we calculated the electric field at a location on the axis of a uniformly charged ring. Without doing all those calculations explain why we canโ€™t use Gauss law to determine the electric field at that location.

The electric field is horizontal and has the values indicated on the surface of cylinder as shown in Figure 21.65. What can you deduce from this pattern of electric field? Include a numerical result.

Figure 21.70 shows a close-up of the central region of a capacitor made of two large metal plates of area , very close together and charged equally and oppositely. There are +Q and -Q on the inner surfaces of the plates and small amounts of charge +q and -q on the outer surfaces.

(a). Knowing Q, determine E: Consider a Gaussian surface in the shape of a shoe box, with one end of area Abox in the interior of the left plate and the other end in the air gap (surface 1 in the diagram). Using only the fact that the electric field is expected to be horizontal everywhere in this region, use Gaussโ€™s law to determine the magnitude of the field in the air gap. Check that your result agrees with our earlier calculations in the Chapter 15. (Be sure to consider the flux on all faces of your Gaussian box.)

The electric field on a closed surface is due to all the charges in the universe, including the charges outside the closed surface. Explain why the total flux nevertheless proportional only to the charges that are inside the surface, with no apparent influence of the charges outside.

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