Quiz 2 Ch. 22 -  2*,6*,5*, 4*,8*,20*,38*, 23*, 54*, 21*, 30,* 44*, 29*, 32*, 45*, 46*, 16 Feel free to print out and start  your work in blanks below  with this page as your  cover sheet. (ANSWERS) 
EVEN ANSWERS
* DISCUSSIONS PROVIDED
HYPERPHYSICS  NOTE: THIS HYPER PHYSICS LINK ,  Gauss's Law,  IS ESPECIALLY USEFUL.
DISCUSSIONS
2 THE DIRECTION BETWEEN THE NORMAL LINE ,  ASSUMED TO POINT UPWARD , AND THE ELECTRIC FIELD IS 70 DEGREES.   SEE FIG. 22.6 
6. Look downward in the negative z direction toward the x-y plane. 

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

Let's look at each side
1: Only the  y-component of the electric field contributes.
3: Again only the y-component contributes.
5: Only the x-component of the electric field contributes:
6: Again only the x-component contributes.
5. This problem is potentially tricky but not necessarily so.  Let  hemisphere have its axis along the z axis and be concave down i.e. the hump points upward along z . Let r  be the  radial vector magnitude measured from the origin to any point on 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)   
4.  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
20. (a) See example 22.5 and 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. 
38. 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 
THE ABOVE DESCRIPTION COULD CORRESPOND  to the line being above the plane. With that assumption, consider the region between  line and plane and below the negative plane. Clearly between the two structures, the net electric field could not be zero in the same way the electric field does not vanish  at a point on the axis between the two equal and opposite charges of an electric  dipole ,  also described here compared to a magnetic "dipole."
23. See example 22.9. 
54.  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 position at  center. 
21.  See example 22.5 and figures 22.23 (a), (b) and (c), especially (b) and (c). See also example 22.11 and figure 22.27 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. 
30. (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) .
44. 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 .   
29. 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. Also note

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. 

NOTE ABOVE WE USED THE ABSOLUTE VALUE OF THE CHARGE DENSITY,  THE POSITIVE CHARGE DENSITY ON LEFT PLATE.
The is equivalent to the "proper " method, assuming  the outward pointing normal lime and the electric field are oppositely directed on  S4's endcap between  plates. In this case, the angle between vector -E and  normal is 180 degrees so we' d have:
E*A*cos 180 = charge density/epsilon, leading to:
-E*A = charge density/epsilon:
 But note the charge density on right plate is negative: charge density =
 -|charge density| on the right plate. If you substitute this negative sign on the right side of above equation, the signs cancel leading to the top stated relationship---E*A = |charge density|/epsilon:

32. See example 21.12  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. A more direct use of Gauss's Law will be discussed in class during review.
45. The shell surrounds the concentric solid conductor. See #44 DISCUSSIONS above. 
46.  See example 22.5 and figures 22.23 (a), (b) and (c), especially (b) and (c). See also example 22.11 and figure 22.27 for extra reinforcement. 
16. FLUX =  E*(4 *pi*r2) = |q enclosed|/epsilon. Note:  r = Mars' radius,  which if known,  allows you to  compute E. 
CHECK BACK FOR MORE DISCUSSIONS