1. (36 pts.) 4B SP12 REFERENCE: SOLID INSULATING SPHERE QUIZ 2, #23, #54

    Ch. 24, Spherical Symmetry, model of proton?:     A solid sphere of total radius R has charge uniformly distributed throughout its volume. The charge density of the sphere is , where < 0 and the units are C/m3= Coulombs/m3. Answer the following questions using symbols R, and other relevant constants. Show all steps !
    (a)(4 POINTS) What is the total charge of the sphere ? Use symbols.
    Next are several smaller questions to get your mind open:
    (b) (1 POINT) What is the magnitude E of the electric field at the center, point A ?

    2. (c) (1 POINT) At point C , what is the direction of the electric field vector ?

    Indicate this direction with an arrow at that point.

    (d ) ( 1 POINT) At point B , what is the direction of the electric field vector ?

    Indicate this direction with an arrow at that point.
    (e) (2 POINTS) Given that point D is at the surface of the sphere, then what is

    the magnitude E of the electric field at point D ? Use symbols.
    (f ) (1 POINT) At point D , what is the direction of the electric field vector ?

    Indicate this direction with an arrow at that point.
    (g) (2 POINTS) Suppose the magnitude of the electric field at point F is one-

    half the magnitude of the electric field at D. What is the distance of point F

    from the center ? Use symbols.
    Now for the "real deal." We're talking 24 points, so proceed

    carefully. Use symbols always:
    (h) (12 POINTS) Suppose the distance of point B from the center is

    R/4. What is the magnitude E of the electric field at point B ?

    (i) (12 POINTS) Suppose the distance of point C from the center

    is 3R/2. What is the magnitude E of the electric field at point C ?

     

  2. (30 pts.) 4B SP12 REFERENCE: POTENTIAL OF A THIN ROD: CONSERVATION OF ENERGY :
    QUIZ 3     #79;#34, #9, #12, #14, #16;

    Now check this out.     A rod of length a has a linear charge density < 0 in units C/m = Coulombs/m. See fig. 1--This might be the model of the positive electrode in a gas discharge tube defining your office's fluorescent light bulb.
    For all solutions below, use symbols.
    (a) (2 points) What is the total charge on the rod?

    3. (b) (13 points) What is the potential V at the point P shown in fig.1 ?
    Derive this. Show all calculus steps !!
    Now look at fig. 2. One of the ions in the tube is a point charge q of mass m. Let's say it starts from rest at a distance 2b from the end of the rod, and accelerates to a distance b/2 from that end. Pretend that gravity has been turned off !
    (c) (3 points) What must be the sign of q , positive or negative?
    (d) (12 points) What is the speed of the point charge q when it reaches the distance b/2 ?

     

     

     

     

  3. EXTRA CREDIT. 9 points.
    A very long, cylindrical capacitor has linear charge density ( in coulombs/m) on the inner shell and linear charge density (same units, different sign !) on the outer shell. The outer shell has radius b and the inner radius a. Check the diagram below. It might be the cable siphoning free signal from your neighbor's line to your own wide screen TV. Here's some of the physics you should know about this illicit scheme. Use symbols always:

(a) (2 points) What is the magnitude E of the electric field at a distance r from the central axis of the capacitor if b < r ? Explain.
(b) (3 points) What is the magnitude E of the electric field at a distance r from the central axis of the capacitor if a < r < b? Derive E. Show all steps.
(c) (4 points) Derive the capacitance per unit length of the system.

Solution: 
(a) E = 0 since zero  net charge is enclosed by any Gaussian surface larger than b in radius.
(b) E = 2·k·lambda/r. You better derive this ! Remember that for cylindrical symmetry, 
(E·2·pi·r·L) = qenc /e0 = lambda·L/e0; thus, E = lamba/( 2·pi·e0.r ) = 2·k·lambda/r. Note qenc =lambda·L, since the radius of the Gaussian surface is between a and b and only encloses lambda·L on the inner sphere !
(c) For a length L of the capacitor, the capacitance is Q/V = (lamba·L)/[2·k·lambda·ln(b/a)] = L/[2·k·ln(b/a)].  Note that V = 2·k·lambda·ln(b/a) is obtained by integrating 2·k·lambda/r  from a to b. To get the capacitance per unit length, divide by L to get  1/[2kln(b/a)] . See the derivation on page 808, example 26.2.