Quiz 2 Ch. 22 -  2**, 4**, 5*,  7*, 8**, 22**, 36**,  25*, 52**, 23*, 28** 42**, 27*, 30**, 43*, 44**, 18** Feel free to print out and start  your work in blanks below  with this page as your  cover sheet.
* DISCUSSIONS PROVIDED/** EVEN ANSWERS PROVIDED
DISCUSSIONS
2. THE DIRECTION BETWEEN THE NORMAL LINE ,  ASSUMED TO POINT UPWARD , AND THE ELECTRIC FIELD IS 70 DEGREES.   SEE FIG. 22.6 
Below is another way of seeing the problem:

The field is uniform and the surface is flat, so use .

Set Up:   is the angle between the normal to the surface and the direction of , so .

Execute: 

Evaluate: If the field were perpendicular to the surface the flux would be  The flux in this problem is much less than this because only the component of perpendicular to the surface contributes to the


4. ASSUME THE NORMAL LINE  ON EACH FACE  POINTS OUTWARD FROM WITHIN THE CLOSED CUBE. First of all, the electric field points along the x-z plane which makes the angle with  the normal lines  on side 1 and side 3  90 degrees-- there is no flux through those sides . Flux only possibly goes through sides 2, 5, 4 and 6 ---the remaining 4 sides since two are removed from the list. 

Let's look at each side
2: Only the  z component of the electric field contributes:
 flux = 3.00 (N/C)*(0.300 m)2* cos 0  
4: Again only the z-component contributes, but the value of the z-component of the field at  z = 0 is ZERO, so the flux through 4 is ZERO.

Use similar  logic to evaluate the flux through sides 6 and 5, where only the x-component  of the field contributes. Hint ~  the flux through side 6 is ZERO. 

Below is another way of seeing the problem:

Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate the total enclosed charge.

Set Up:  . The area of each face is , where .

Execute:  .

. .

.

(b)Total flux:   Therefore,

Evaluate: Flux is positive when is directed out of the volume and negative when it is directed into the volume.


5. This problem is potentially tricky but not necessarily so if you take care.  Let the hemisphere have its axis along the z axis and be concave down i.e. the hump points upward along z . Let r  be the magnitude of the radial vector from the origin to any point on the hemisphere. Go here http://cnx.org/content/m13600/latest/  and scroll down to the section on spherical coordinates to a  nice model of the problem at hand.  Let the electric field point in the positive z direction. Think about a  differential surface that is a special circular strip.  The radius is*r*sin phi so the circumference is  
 2*pi*r* sin phi, where phi is the angle the radial vector   makes with the z axis ,  also called the polar angle.  The differential thickness of the strip is r*d phi.  Thus the differential area is 
2*pi*r2*sin phi * d phi, where the range of integration will be from phi = 0 to phi = pi/2 (90 degrees) SINCE WE HAVE A HEMISPHERE.  The electric field  vector at each point on the surface makes an angle phi with the differential  section of surface  .  Note we have integrated around the full range --2*pi--of the azimuthal angle theta---explained at  http://cnx.org/content/m13600/latest/  --- which is why the radius of the strip was designated 2*pi*r sin phi, where the 2*pi reflects the procedure around a full (azimuthal) circle.    Thus 
E*dA*cos phi = E*2*pi*r2*sin phi*cos phi * d phi.
Integrate this expression over phi  between 0 and pi/2. USE TRIG IDENTITY---sin phi * cos phi = (1/2)*sin 2 phi. 

Note you can get the same result by realizing the  flux through the hemispherical surface  is the same---IN MAGNITUDE ONLY---as the flux through the flat circular surface of radius r bounded by the circular   hemisphere bottom edge. In this case the flux is E*(area of the flat circle of radius r)   
7.  See example 22.6,  reviewed in class. Note the flux   through the "curvy " side, which has area proportional to r,  is independent of radius r using the derived  formula  for the electric field magnitude which varies as  1/r. So the two powers  of r cancel. 
8.  In each case the integral of E*dA* cos theta = charge enclosed/epsilono  

Below is another way of seeing the problem:

Apply Gauss’s law to each surface.

Set Up:  is the algebraic sum of the charges enclosed by each surface. Flux out of the volume is positive and flux into the enclosed volume is negative.

Execute: (a)

(b)

(c)

(d)

(e)

Evaluate: (f) All that matters for Gauss’s law is the total amount of charge enclosed by the surface, not its distribution within the surface.


22. (a) See example 22.5 and figure figure 22.17 for any general conductor.  The charged spherical conductor's radius is R = 0.100 m, so any values of r < 0.100 m will produce zero electric field.
(b) See example  22.6 and  figure 22.17 for any general conductor,  including a long thin line as in this part.  The charged linear conductor's radius is R = 0.100 m, so any values of r < 0.100 m will produce zero electric field. 
(c) The electric field is constant on a given side of the sheet. See example 22.7. 

Below is another way of seeing the problem:

Apply the results of Examples 22.5, 22.6 and 22.7.

Set Up: Gauss’s law can be used to show that the field outside a long conducting cylinder is the same as for a line of charge along the axis of the cylinder.

Execute: (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

36. Let the charged long line  and the plane of charge both be perpendicular to the page you are looking at. Let the plane be be horizontal on your page.  Examine any point directly  vertically above  the line of charge. In this case, the  electric field due to the line of charge points up and the field from the plane points down. Find the distance above the line where the two fields cancel using formulae for the field magnitude E in example 22.6 and example 22.7 
Below is another way of seeing the problem:

Identify: The  particle feels no force where the net electric field due to the two distributions of charge is zero.

Set Up: The fields can cancel only in the regions A and B shown in Figure 22.36, because only in these two regions are the two fields in opposite directions.

Execute:  gives and .

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

Evaluate: 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 22.36


25. See example 22.9. 
52.  More discussions  on this will follow , but the starting point of this historical problem is  example 22.9.  Find the electric field as a function of r inside the NON-conducting sphere AND  then find the force on an electron inside the sphere. That force should always point to the center of the sphere and will support simple harmonic oscillations for small displacements of the electron from the equilibrium positron at  center. 

Below is another way of seeing the problem:

Identify: Example 22.9 gives the expression for the electric field both inside and outside a uniformly charged sphere. Use to calculate the force on the electron.

Set Up: The sphere has charge .

Execute: (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.

Evaluate: As long as the initial displacement is less than R the frequency of the motion is independent of the initial displacement.


23.  See example 22.5 and figures 22.23 (a), (b) and (c), especially (b) and (c). See also example 22.11 and figure 22.28 for extra reinforcement.

(a) Before you inserted the point charge in the cavity, the problem looked like  fig. 22.23 (b); after you insert the charge inside the cavity, we have fig. 22.23 (c). READ EXAMPLE 22.11 IN DETAIL. Find the charge on the outer surface as in example 22.11; to get  charge density,  compute density =
(outer surface charge)/(outer surface area), where the surface area of any sphere is 4*pi*r2. The outer radius is given.
(b)  The electric field on the outside surface of the conductor has magnitude E  given by  equation 22.10; thus you need the charge density on the outer surface given in part (a). Again, that density = 
 (outside surface charge)/(outer surface area)
(c) The flux = (charge enclosed)/epsilon. Charge enclosed =  -0.500x10-6 C. 
28. (a) At this location the sheet appears to be infinite and you would use equation E = (surface  density)/(2*epsilon) as in example 22.7. 
(b) At this distance the sheet would behave  like a point charge; thus E = k*Q/D2, where D = 100 m and Q = (charge density)*(sheet area) .

Below is another way of seeing the problem:

Identify: Close to a finite sheet the field is the same as for an infinite sheet. Very far from a finite sheet the field is that of a point charge.

Set Up: For an infinite sheet, . For a point charge, .

Execute: (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.

Evaluate: The sheet can be treated as infinite at points where the distance to the sheet is much less than the distance to the edge of the sheet. The sheet can be treated as a point charge at points for which the distance to the sheet is much greater than the dimensions of the sheet.

 

 

 
42. See  figures 22.23 (a), (b) and (c), especially (b) and (c). READ EXAMPLE 22.11 IN DETAIL. You can use Gauss's Law to explain  part (a). 
(a) Clearly E = 0 for  b < r < c since this double inequality (i.e. range) refers to  points  within the hollow conductor . For  r < a, we also have E = 0. For  a < r < b,  we have the electric field due to a point charge of magnitude | q|. i.e. ,  magnitude E = k*q/r2.     For r > c, we have for the flux E*(4 *pi*r2) = (q enclosed)/epsilon, where r > c and q enclosed = q. 
(b) The graph of E is described in the previous  part. 
(c) See example 22.11 and treat the conductor of radius a  within the cavity as a point charge when you draw a Gaussian spherical surface  of radius  r concentric with the system with  a < r < b. 
(d)  See example 22.11.
(e) The field lines will be radial and disappear inside the conductors. They begin on the surface of the small conductor of radius a inside the cavity. They are radial inside the space   a < r < b and disappear for   b < r < c, terminating on the inner surface of the hollow conductor,   and reappearing  radially when c < r outside the hollow spherical  conductor .   

Below is another way of seeing the problem:

Identify: Apply Gauss’s law.

Set Up: Use a Gaussian surface that is a sphere of radius r and that is concentric with the conducting spheres.

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

For  since there is +q inside a radius r.

For 0, since since these points are within the conducting material

For  since again the total charge enclosed is +q.

(b) The graph of E versus r is sketched in Figure 22.42a.

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

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

(e) The field lines are sketched in Figure 22.42b. 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 .

                                

Figure 22.42


27. Assume negligible charge on the outside surfaces of the plates. We use S2 and S3 to find the field outside the plates and since no charge is enclosed within these Gaussian surfaces, the field outside must be ZERO.

Now with regard to S4, we have a double argument which I'll explain now:
Clearly on the Gaussian surface, E*A = (charge density)*A /epsilon, where A is the areas of the end cap between the plates. Thus E = 
(charge density /epsilon, E = 0 on the other end cap within the right conducting plate of finite thickness. NOTE IN THE CASE OF S4, E IS THE MAGNITUDE  of the electric field due to both the positive and negative surfaces charges on the surfaces facing each other across the gap.

You can use a second argument to get the  electric field of magnitude E between the plates. Assume the oppositely charged inner surfaces can be represented as infinite charged parallel  non-conducting sheets. The net field would be the sum of the field of the two sheets E =
(charge density /(2epsilon) + charge density /(2epsilon) = 
charge density /epsilon, same as above. 

30. See example 21.13  for help on using superposition to find the net field by adding the individual fields  for each sheet. This problem is a little more  complicated  because we have 4 sheets but same principles apply.

Below is another way of seeing the problem:

Set Up: The electric field of a large sheet of charge is . The field is directed away from a positive sheet and toward a negative sheet.

Execute: (a) At .

(b)  .

(c) .

Evaluate: The field at C is not zero. The pieces of plastic are not conductors.

43. The shell surrounds the concentric solid conductor. See #42 DISCUSSIONS above. 
44.  See example 22.5 and figures 22.23 (a), (b) and (c), especially (b) and (c). See also example 22.11 and figure 22.28 for extra reinforcement. ALSO SEE #23 DISCUSSION.

Below is another way of seeing the problem:

Identify: Apply Gauss’s law and conservation of charge.

Set Up: Use a Gaussian surface that is a sphere of radius r and that has the point charge at its center.

Execute: (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 –2Q.

(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 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 22.44a.

(e) The graph of E versus r is sketched in Figure 22.44b.

                              

Figure 22.44

Evaluate:  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.


18. FLUX =  E*(4 *pi*r2) = |q enclosed|/epsilon. Note:  r = Mars' radius,  which if known,  allows you to  compute E. 

Below is another way of seeing the problem:

Identify: According to Exercise 21.32, the Earth’s electric field points towards its center. Since Mars’s electric field is similar to that of Earth, we assume it points toward the center of Mars. Therefore the charge on Mars must be negative. We use Gauss’s law to relate the electric flux to the charge causing it.

Set Up: Gauss’s law is  and the electric flux is .

Execute: (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

Evaluate: Even though the charge on Mars is very large, it is spread over a large area, giving a small surface charge density.


CHECK BACK FOR MORE DISCUSSIONS