QUIZ 11 , CHAPTER  30 PROBLEMS :
5, 6, 7, 9, 15, 16, 17, 22, 24, 31, 32, 39 , 40,  48, 49; Do 48 and 49 for deep understanding  after doing the following: 54, 55.
Turn in only: 6, 16, 22, 24, 32,  54

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

 6. See example 30-3.  The formula for L is derived and used to solve a  similar numerical problem. 
 16. Use formula 30-7 and formula 24-6. While I did not derive the former equation , it follows analogously from the derivation of 24-6, which I did review  for sure.  For part (b),  set the energy densities equal, writing  uE = uB  and solve for the electric field magnitude E. 
22.  Use formula 30-9 for  computational problems which I reviewed in class:  For  (a),  set 0.95 = (1 - e-Rt/L) and solve for the time using manipulation and by taking the natural log (ln) of both sides of the equation. Similar methods apply to (b) and  (c). 
32. Yo, we did this in class. We said the picture would look different than figure 30-11 in that the charge would initially be zero and the current would initially be at maximum. In other  words we said Q = -Qmsinwt because the current would be  draining the plate when the charge on it is zero.  That's  a recipe for an initially decreasing charge,  i.e. a negative sine function. To get the current, just differentiate as follows: I = -dQ/dt. The negative sign is critical since the charge  decreases as the positive current exits  plate.

Note  the charge amplitude Qm is the maximum charge.  The maximum charge  can be found by equating the
(total energy when the current is zero) to (the total energy when the charge is zero). (1/2)*Qm2/C=(1/2)*L*Io2. Solve for the maximum charge in terms of Io and insert it into the formula for the charge Q vs t.