ANSWERS

Quiz 2  Chapter 21 problems: QUIZ 1, CHAPTER 21: 46*, 47*, 48* (DIPOLE PRACTICE), 49*, 59*, 62*, 65*, 80*, 86*, 91*
*  means discussions provided. DUE DATE : WEDNESDAY SEPTEMBER 8.  MASTERINGPHYSICS ASSIGNMENT IS  POSTED. 
Below is a link to a discussion of Gauss's Law (Ch.22, Quiz 3 TBA)  presented by Richard P. Feynman.
( http://en.wikipedia.org/wiki/Richard_Feynman  ).  He shows there is no position of stable equilibrium 
in a configuration of charges in space. Atoms are stable only because the electrons are held in "orbits" according to 
Quantum Mechanics and have well defined energy states. The positive nucleus pulls the electrons in close but there is a
spread in the possible  locations of the electrons due to the Heisenberg Uncertainty Principle--if the electrons were precisely localized, there would be a large uncertainty in the electrons' momentum, resulting in some electrons with large enough momentum and energy  escaping the electro-static grip of the nucleus. Since such escapes are not observed, the electrons must be "spread out" within "clouds."   

More practically, through Chapter 22 knowledge you can compute more easily fields using symmetry. For example you can find  the field near a long line of charge faster than in Chapter 21 where computations are more tricky.  In Ch. 21, the long line of charge  calculation is done for you and is related to results for problems 46 and 47 below.  

http://www.scribd.com/doc/20697866/Vol-2-Ch-05-Application-of-Gauss-Law

As Feynman above and our own textbook observe, one consequence of Gauss's Law is the formula for the magnitude of the perpendicular electric field on the surface of a charged conductor (in equilibrium) : E= (charge density)/epsilon = (charge density)/eo . If a conductor tapers to a sharp point, the charge density will be very high in that region, resulting in a large electric field--large enough to "pull" electrons from the chemical grip of the surface, assuming the tip has excess negative charge on the surface.   The electrons move along the field lines through a vacuumed space and strike a positive screen. Electrons recorded as having a lower speed were pulled from conductor sections with strong chemical binding and those higher speed from regions of low chemical binding. From such information we get a map of the chemical surface structure on  the electron emitting,  tapered conductor: http://en.wikipedia.org/wiki/Field_emission_microscopy . 

Feynman observes that Quantum Mechanical effects limit the field emission microscope's resolution. (Page 6-14, Vol. 2, The Feynman Lectures on Physics.) Since the electrons cannot be precisely located without a  significant uncertainty in momentum and energy, there is a spread in the location of  electrons striking the screen, producing a smearing of spots.  To solve this problem, scientists and engineers,  who took the equivalent "4B" in their day, invented the field-ion microscope,  which employs the motion of heavier ions instead of electrons  toward  the screen. The field-ion microscope is reverse polarized, so the tip is positive and the  screen in now   negative in contrast to the electron field emission device.  Since He ions, for example,  have a much larger mass than electrons, Quantum Mechanics predicts a much smaller wavelength for the "matter waves" defining them.  Hence, there is better resolution on the screen as you will learn in more detail next term.  (Here's how it works.  When the Helium atoms in the bulb area come close to the positive  microscope tip, the powerful electric field strips an electron off the atoms, turning them into  positive ions. An ion then moves along a field line toward the negative screen. The ionization and energy of the Helium atom   varies over the region of the tip,  made of Tungsten for example.  Thus scientists can  observe a picture of the configuration of atoms in Tungsten. In particular,  the rate of Helium ionization at the location of a Tungsten atom is different than in the space between atoms, creating a image of their arrangement. ( http://en.wikipedia.org/wiki/Field_ion_microscope. )

After a recent lab, some intense discussion occurred  dealing with the dynamics toward equilibrium of charges on the surface of a conductor and why charges  tend to concentrate at sharp corners--such as a microscope tip.  One can observe this dynamic in the case of a sphere which is later deformed into a ellipsoid.  First we show the sphere with eight charges:

We focus on the positive charge adjacent and to the  left  of  the one at the very top of the sphere. The lines from the other charges converge on our charge of choice.  It should be clear to you that the lines are in the same direction as the forces on our charge of choice and  the net force on that charge is perpendicular to and outward from the sphere at that point. Thus, so they don't  move, the surface charges on a spherical conductor  must be evenly distributed to ensure each charge experiences a force perpendicular to the surface.

Now deform (flatten) the above sphere into an  extreme ellipsoid with major axis parallel to the horizontal of this page  but for the moment maintain the same separation distances as on the sphere. Remember, charges have mass and thus inertia, so  there would  be a  time lag  between the  deformation and when the charges  completed redistributing themselves on the surface.  Will the  net force on our charge of choice  be perpendicular (normal)   to the surface?  Or will  our charge of choice  tend to move leftward  until the net force it experiences is perpendicular to the surface ,  suggesting charges have a smaller separation distance and higher concentration  in the regions centered at either horizontal apex?   

Our  post lab discussion perhaps suggests  the above picture showing some of the surface charges at equilibrium. The normal  and  tangent to the surface are shown  at a point in  the region near the apex. 

Simulation: Click here for the motion of an electron in a uniform electric field.

46. See the integral in example 21-11,  just realize you don't integrate from -pi/2 to +pi/2, where pi = 3.14....  .  Instead you will evaluate sin theta at different limits. Have a close look at figure 21-29. Suppose the angle was indeed the maximum angle, and thus the value of y was L/2.  We see that sin theta = opposite/ hypotenuse = L/[2*(L2/4 + x2)1/2 ] . Substitute  sin theta into the upper and lower limits of the integral  at the bottom of page 573.  
47. This is easy if you look at the figure as a pyramid with   four inclined sides converging on the observation point at height z. The electric field due to a given side  makes  an angle  with the horizontal given by  tan theta = z/ (L/2) = 2*z/L, where L is the square's length.  This field has  two components,  horizontal and vertical. When you sum the fields from all 4 sides, the horizontal components  cancel from  symmetry. The 4 vertical components survive; the net vertical component is Ez = 4*E*sin theta. You can identify sine theta by looking at the figure. Hint: Compute the hypotenuse of a right triangle with sides  z and L/2 . Note  E is the  electric field magnitude at a point in space a  distance (L2/4 + z2)1/2 from a side. See problem 46.  
48. See the derivation on page 580. It's the same problem if you make the substitution L/2 = a and r = x. Same problem, different symbols.  However you should understand the problem's essence not just be able to "reverse-engineer'" through clever substitutions.  Also give the direction of the field. Does it point vertically up or vertically down  the page?  
49. This is an easy problem if you realize  from symmetry the field must point horizontally along the x -axis in the negative direction (left). The leftward differential field has x-component   dEx = -[k*Lambda*dL/R2]* cos @,  where angle @ ranges from -@o  to + @o.  Note that dL = R*d@.  Integrate  -[k*Lambda*R*d@/R2]* cos @ = 
 -k*Lambda cos@ d@ /R between -@o  to + @o .  

59.  This problem is  identical to the future lab  dealing with an electron moving  in a uniform electric field.  Click here . The exit angle theta is given by tan theta = |Vy |/ |Vx|, where  Vy and Vx will be   covered in  pre-lab discussions. 
62.  See class references  and section 21-11. The torque and potential energy are derived and discussed in detail using comparisons with  simple harmonic motion of a mass attached to a  spring. Use equations 21-9 and 21-10.
65. 
(a) As discussed in class the condition for simple harmonic motion was  angle @ being so small  you could say
 sin @ = @ as a first order approximation. With the small angle approximation  you write, 

 I*d2@/dt2 = -p*E*@, where p is the electric dipole moment magnitude,  I is the moment of inertia about the charges'  midpoint  and E is the electric field magnitude.  

(b) See class notes.  
80.  Review equilibrium from Physics 4A.  Consider for example the right sphere of charge Q/2 and mass m = 0.0024 kg.
There are three forces acting  on the hanging mass: the tension force directed along the wire at   26 degrees with the vertical, the down ward gravitational force, and the electric force  directed horizontally right. Break each force into horizontal and vertical components. Note the gravitational force, aka weight,  points  downward. Only the tension force of magnitude T has two non-zero components.
 
For the x direction:

T*sin26 = kQ2/(4d2 ), where d is the horizontal distance between the two charges. You can easily find distance d  by inspection and  trigonometry applied to the right triangle of hypotenuse.0.78 m 

The second equation is for the y-direction:

T*cos26 =  mg.

Solve the above  x and y-direction equations simultaneously for Q and T. 
86. This is an easy problem if you know the centripetal force keeps the  electron in orbit  about the wire's center. The centripetal force in this case  is the electric force  directed toward the wire. That force has magnitude e*E, where e and E are the electric charge and electric field magnitudes, respectively.  To find E, which is a function of r,  see example 22-6 or example 21-11, which give similar results through different means. Note that in example 21-11,  x = r and in example 22-6, R = r. 

Now, we know that mv2/r = e*E, where E is inversely proportional to r  as shown in the above-mentioned examples.
(a) See examples 22-6 and 21-11.
(b)  Use mv2/r = e*E. Is the value of r needed to solve this problem? Explain.
91. This problem is similar to #80. The hanging mass is in equilibrium. Now the question is whether the plane is positive or negative. If the plane is negative, then the electric force is down, in which case the tension would be more  than the weight  in magnitude and 0 = T - Q*E - mg. If the plate was positive, the electric force would be directed up, in which case the tension would be less than the weight in magnitude and  0 = T + Q*E - mg. You must make your choice by comparing T with mg.

Above,  T is the tension force magnitude, Q is the charge,  E is the electric field magnitude and mg is the weight. Mass m, charge  Q and tension magnitude T are given.   

(a) Solve for E.
(b) Solve for the charge density using example  22-7 or example 21-12 in the limit of infinite R (equation 21-7.).