Quiz 2 Ch. 22 ~  Quiz 2  -  2*,6*,5*, 4*,8*,20*,38*, 23*, 54*, 21*, 30,* 44*, 29*, 32*, 45*, 46*, 16 .    
2.
6. 

 (a)  

 

 

 

 

 

(b) The total flux through the cube must be zero; any flux entering the cube must also leave it, since the field is uniform. Our calculation gives the result; the sum of the fluxes calculated in part (a)is zero.
4.
(a)

(b) In the calculation in part (a) the radius r of the cylinder divided out, so the flux remains the same,

(c)  (twice the flux calculated in parts (a)
and (b)).

8.

(a)

(b)

(c)

(d)

(e)


20

(a) For points outside a uniform spherical charge distribution, all the charge can be considered to be concentrated at the center of the sphere. The field outside the sphere is thus inversely proportional to the square of the distance from the center. In this case,

(b) For points outside a long cylindrically symmetrical charge distribution, the field is identical to that of a long line of charge:  that is, inversely proportional to the distance from the axis of the cylinder. In this case

(c) The field of an infinite sheet of charge is  i.e., it is independent of the distance from the sheet. Thus in this case

38.

The fields can cancel only in the regions A and B shown in Figure.38 below, because only in these two regions are the two fields in opposite directions.

     gives  and

The fields cancel 16 cm from the line in regions A and B.

   The result is independent of the distance between the line and the sheet. The electric field of an infinite sheet of charge is uniform, independent of the distance from the sheet.

 

Figure 38

54. (a)
Only at  is  for the uniformly charged sphere.
(b) At points inside the sphere,  The field is radially outward.  The minus sign denotes that  is radially inward. For simple harmonic motion,  where    so  and

(c) If  then  

The atom radius in this model is the correct order of magnitude.

(d) If    and  

The electron would still oscillate because the force is directed toward the equilibrium position at  But the motion would not be simple harmonic, since  is proportional to  and simple harmonic motion requires that the restoring force be proportional to the displacement from equilibrium.

30.

(a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so

(b) At a distance of 100 m from the center, the sheet looks like a point, so:

(c) There would be no difference if the sheet was a conductor. The charge would automatically spread out evenly over both faces, giving it half the charge density on either face as the insulator but the same electric field. Far away, they both look like points with the same charge.

44.

(a) For  since these points are within the conducting material.

For  since there is  inside a radius r.

For  since these points are within the conducting material.

For  since again the total charge enclosed is

(b) The graph of E versus r is sketched in Figure 44a.

(c) Since the Gaussian sphere of radius r, for  must enclose zero net charge, the charge on the inner shell surface is

(d) Since the hollow sphere has no net charge and has charge  on its inner surface, the charge on the outer shell surface is

(e) The field lines are sketched in Figure 44b. Where the field is nonzero, it is radially outward.

Evaluate:   The net charge on the inner solid conducting sphere is on the surface of that sphere. The presence of the hollow sphere does not affect the electric field in the region

 

32.

(a) At

(b)

(c)  

46.:   (a) For      radially outward, since the charge enclosed is Q, the charge of the point charge. For    since these points are within the conducting material. For    radially inward, since the total enclosed charge is

(b) Since a Gaussian surface with radius r, for  must enclose zero net charge, the total charge on the inner surface is  and the surface charge density on the inner surface is

(c) Since the net charge on the shell is  and there is  on the inner surface, there must be  on the outer surface. The surface charge density on the outer surface is

(d) The field lines and the locations of the charges are sketched in Figure 46a.

(e) The graph of E versus r is sketched in Figure 46b.

 

   

Figure 46

 

For  the electric field is due solely to the point charge Q. For  the electric field is due to the charge  that is on the outer surface of the shell.

16.

: (a) Solving Gauss’s law for q, putting in the numbers, and recalling that q is negative, gives

(b) Use the definition of electric flux to find the electric field. The area to use is the surface area of Mars.

(c) The surface charge density on Mars is therefore