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Chapter 21: Q2Q (page 897)

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 can’t use Gauss law to determine the electric field at that location.

Short Answer

Expert verified

The electric field at the location on the axis of uniformly charged ring because field lines are not always perpendicular to the enclosed area for that location on the axis of ring.

Step by step solution

01

Conceptual Explanation

The Gauss law is used to find the electric field inside an enclosed surface by equating net flux inside the enclosed surface to net charge divided by permeability of free space.

02

Necessary condition to apply Gauss Law

The Gauss law can be only applicable for the surfaces in which the electric field lines are always perpendicular to the surface area. This condition is fulfilled only in symmetrical objects like sphere, cylinder etc.

03

Reason for avoiding the use of Gauss law to find the electric filed at a location on the axis of uniformly charged ring

The electric field lines from the uniformly charged ring are not always perpendicular to the enclosed area in which the location to find electric filed lies so Gauss Law is not applicable to find electric field at location on the axis of the uniformly charged ring.

Therefore, the electric field at the location on the axis of uniformly charged ring because field lines are not always perpendicular to the enclosed area for that location on the axis of ring.

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Most popular questions from this chapter

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 Gauss’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 center of a thin paper cube in outer space is located at the origin. Each edge is 10 cm long. The only other objects in the neighborhood are some small charged particles whose charges and position at this instant are following: +9 nC at (4,2,-3) cm, -6 nC at (1,-3,2) cm, +7 nC at (15,0,4) cm, +4 nC at (-1,3,2) cm and -3 nC at (-2,12,-3) cm. The electric flux on the five of the six faces of the cube totals 564 V m. What is the flux on the other face?

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?)

The electric field has been measured to be vertically upward everywhere on the surface of a box 30 cm long, 4 cm high and 3 cm deep as shown in Figure 21.64. All over the bottom of the box E1=1500V/m , all over the sides E2=1000V/mand all over the top E3=600V/m. What can you conclude about the contents of the box. Include a numerical result.

In Figure 21.15 the magnitude of the electric field is 1000 V/m, and the field is at an angle of 30oto the outward-going normal. What is the flux on the small rectangle whose dimensions are 1 mmby2 mm ?
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