ANSWERS

Quiz  3 Chapter  22 problems: 4*, 9*; Spherical symmetry -2*,  13*, 14*, 16*, 18*, 19*, 21*, 22*; Planar symmetry-24*, 26*, 46*; cylindrical symmetry-15*, 35*, 36

* discussions provided below or in class

Below is a link to a discussion of Gauss's Law (Ch.22, Quiz 3)  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. )

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

DISCUSSIONS:
2. Flux helps you find the electric field magnitude  E which is positive. You integrate E*dA*cos theta over the entire spherical Gaussian surface. Now recall the normal line always points out from the closed surface. On the other hand you know that the electric field will point toward the center of the negative spherical Earth.  Thus the angle  between the electric field and the normal line is 180 degrees. Thus: E*dA*cos theta = E*dA* cos 180 = E*dA*(-1) = - E*dA.  When you integrate,  you get net flux = 
-E*(surface area ), where surface area is the surface area of a sphere with the Earth's radius. Thus the flux will be negative.    Note: E = 150 N/C. 
4. 
(a) The flux through the curved part is the same as the flux through an imaginary flat circular part bordered by the hemi-sphere's "rim." The hemi-sphere is an open surface; the normal is perpendicular to the curved part and points outward inclined toward the right, changing direction over the hemisphere. The normal line through the imaginary flat part would point horizontally to the right in the same direction as the electric field vector. Thus the flux is E*A*cos 0 = E*A, where A is the area of a circle of radius r and E is the *magnitude* of the electric field. Be careful. 

Note: We use the following argument to deduce the flux through the imaginary flat disk is the same as that through the hemisphere. Pretend the surface was closed, i.e. we "cap" the left rim with a flat, circular disk. The normal through the flat disk part would thus point outward from the interior of the surface, toward the left. Since no charge is enclosed, the integral over the entire closed surface would be zero, thus making the flux through the disk the negative of the flux through the hemisphere, i. e. flux disk + flux hemisphere = 0 so that flux disk = - flux hemisphere if the surface was closed.  In this problem, 4, the surface is open and we want to find the flux through the hemispherical surface. From the above arguments for a closed surface with no charge inside , | flux hemisphere | = | flux disk | = E*(area of the circle) . 

(b) In this case the normal line through the flat part and the electric field would make an angle of 90 degrees. Hint: What is cos 90 ? 
9. Net flux = EL*A*cos 180 + ER*A*cos0, where A is the area of a square of side 25 (m). EL is the magnitude of the field at the left face and ER is the magnitude of the field at the  right face. Note that the normal line points outward from this closed surface (box).  Set the net flux = (charge enclosed)/eo.  Solve for (charge enclosed).

 13. Flux helps you find the electric field magnitude  E which is positive. You integrate E*dA*cos theta over the entire spherical Gaussian surface. Now recall the normal line always points out from the closed surface. On the other hand you know that the electric field will point toward  the center of the spherical conductor with excess negative  charge on the surface.  Thus the angle  between the electric field and the normal line is 180 degrees. Thus: E*dA*cos theta = 
E*dA* cos 180 = E*dA*(-1) = - E*dA.  When you integrate,  you get a net negative flux: -E*A, where A is the surface area of the sphere.    Note: E is given.  

Set the net flux = (charge enclosed)/eo.  Solve for (charge enclosed).
16. Flux helps you find the electric field magnitude  E which is positive. You integrate E*dA*cos theta over the entire spherical Gaussian surface. Now recall the normal line always points out from the closed surface. On the other hand you know that the electric field will point toward  the center of the spherical conductor with excess negative  charge on the surface.  ( For this charge, see little hint below.) Thus the angle  between the electric field and the normal line is 180 degrees. Thus: E*dA*cos theta = E*dA* cos 180 = E*dA*(-1) = - E*dA.  When you integrate,  you get a net negative flux: -E*A, where A is the surface area of the sphere.     

Set the net flux = (charge enclosed)/eo.  Note:  (charge enclosed) is given and is negative. Hint: add the two values of the given charges  to get the net negative charge on the surface. 

Compute the magnitude E of the electric field and draw the electric field lines that point radially toward the sphere's center. 

14.  Flux helps you find the electric field magnitude  E which is positive. You integrate E*dA*cos theta over the entire spherical Gaussian surface. Now recall the normal line always points out from the closed surface. Assume that the electric field  points away the center of the spherical conductor with excess positive  charge on the surface.  Thus the angle  between the electric field and the normal line is 0 degrees. Thus: E*dA*cos theta = E*dA* cos 0 = E*dA*(1) = 
 E*dA.  When you integrate,  you get a net positive flux = E*A, where A is the surface area of the sphere.  

 Set the net flux = (charge enclosed)/eo.  Now the charge enclosed by the Gaussian surface (just above the sphere surface)  may be written as Q that is  uniformly distributed on the sphere surface. Thus: E*A = Q/ eo . When you divide both sides by A you get: E = (1/eo)*(Q/A) which leads the result you seek. Q/A is the symbolically given  surface charge density in (C/m2).  A is the surface area of the sphere. Thus the field on the surface of a charged spherical conducting shell is the same as the electric field on a charged solid conducting sphere. In both cases, the electric field below the surface is zero.
18.
(a) What is E inside a conductor in equilibrium?
(b) What is E inside a conductor in equilibrium?
(c) See #16. Flux helps you find the electric field magnitude  E which is positive. You integrate E*dA*cos theta over the entire spherical Gaussian surface outside the conductor's surface. Now recall the normal line always points out from the closed surface. On the other hand you know that the electric field will point toward the center of the spherical conductor with excess negative  charge on the surface.  ( For this charge, see little hint below.) Thus the angle  between the electric field and the normal line is 180 degrees. Thus: E*dA*cos theta = E*dA* cos 180 = E*dA*(-1) = - E*dA.  When you integrate,  you get a net negative flux: -E*A, where A is the surface area of the sphere.    

Set the net flux = (charge enclosed)/eo.  Note:  (charge enclosed) is given and is negative.

Compute the magnitude E of the electric field and draw the electric field lines that point radially toward the sphere's center.

(d) Same as part (c), but the Gaussian sphere has a larger radius.
(e)  The results should be the same. Why?
(f) See Example 22-4, which will be reviewed. Use the result inside and outside the solid non-conducting sphere as needed for each radius given in parts (a), (b) , (c) and (d).

19. See Example 22-4, which will be reviewed. Use the result inside and outside the solid non-conducting sphere as needed for your plot of E vs r.
35. Try 36 first , then come back to 35 and make appropriate adjustments. Part (d) provides a stimulating  twist: The motion of a charged particle undergoing uniform circular  motion like in # 86, Ch. 21.  As we stated "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. Focus on example 22-6. The field magnitude will have the form 2*k*|lambda|/r , where lambda is given by Q/L and L is the given length. Now, we know that mv2/r = e*E, where E is inversely proportional to r  as shown in the above-mentioned examples. You are given the  value of r = arithmetic mean of the two given radii.
To get the kinetic energy,  manipulate the above centripetal force equation to get (1/2)*mv2  in terms of given symbols.
36. For all of these parts you'd have the following relation for cylindrical symmetry:
E*(2*pi*r*L)*cos theta = (lambda)*L/e0,  where theta = 0 or 180 as the case may be; if lambda for the charge enclosed is negative,  use theta = 180. Otherwise,  if lambda is positive  use theta = 0. Note : pi = 3.14... using your calculator button.  See #35. 
(a) E = 0 since any Gaussian surface at this radius would enclose zero charge. 
(b) E*(2*pi*r*L)*cos 180 = (lambda1)*L/e0   . Here lambda1 is negative and can be found be dividing the total inner radius charge by the cylinder length.
(c) E*(2*pi*r*L)*cos 0 = (lambda2)*L/e0   . Here lambda2 is positive and can be found be dividing the total outer radius charge by the cylinder length.
(d) E*(2*pi*r*L)*cos theta = (lambda1 + lambd2)*L/e0  .  Evaluate  the sum (lambda1 + lambd2) . If  sum  is positive , use theta = 0 otherwise theta = 180.
46.  Will be discussed further in class but here is an outline. 
TO GET THE FIELD INSIDE THE SLAB: Form a cylindrical, if you will, Gaussian surface inside the slab of charge. (You could also use a rectangular Gaussian surface.) Let the cylinder axis be parallel to the x-axis shown in figure 22-40.  See fig. 22-16 for an image of the Gaussian surface in a different context. The surface integral will produce the result 2*E*A =  (charge enclosed)/epsilon, where E = field magnitude at a distance | x | < d/2 from the center, and x is the position. This is true whether x is positive or negative. In this case, -d/2 < x < d/2. To get the field magnitude E, write 2*E*A = 
2*(charge density)*A*| x | and solve for E, assuming  -d/2 < x < d/2, i.e. | x | < d/2. Now let's think about getting the signed x-component of the field Ex in this range, where x represents a subscript. We have 2*Ex = 2*(charge density)* x/epsilon . That formula is consistent with the magnitude and direction of the field physically. When x is negative Ex is negative; when x is positive, Ex is positive. Note: Ex is zero at the origin, suggesting a negatively charged point particle placed there and slightly displaced would undergo simple harmonic motion about x = 0. What would be the frequency if the charges' mass was m and the charge value -q, where q is a positive number? Use these and those of the book as symbols. To find the field outside the slab ie, for x < -d/2 and x > d/2, we write 2*E*A = 2*(charge density)*A*(d/2). This is just the previous formula evaluated at| x |= d/2. Does E vary with distance at points outside the slab?
More discussions later !