Thanks for your reply. Unfortunately I don't know how to proceed. All I can think of is silly: take the derivative of final KE (step 1 above) wrt the radius, and integrate from r=0 to r=0.53. This accomplishes nothing, of course, and results in the same wrong answer I got before.
Homework Statement
Estimate the total path length traveled by a deuteron in a cyclotron of radius 53 cm and operating frequency 12 MHz during the entire acceleration process. Accelerating potential between dees is 80 kV.
mass m = 3.2e-27 kg
charge q = 1.6e-19 C
radius r = 0.53 m
frequency f =...
Thanks for your replies, jtbell and ehild.
Looking at ehild's diagrams, I wonder if my understanding is correct. Imagine a Gaussian pillbox of unit area on the right side of the charged sheet, the box enclosing just the surface charges on that side and extending only halfway into the sheet. The...
I also have Griffiths 3rd. On p. 74 he gives the electric field of an infinite plane with surface charge density sigma. This is (sigma/2e0)n.
On p. 89 he gives the difference in the fields between the upper and lower surface of charge sigma. This is (sigma/e0)n.
I see now that I misquoted...
PS--My confusion increases the more I look at Griffiths (Electrodynamics 3rd edition). On p. 89 note 6, while discussing the field of a sheet of charge, Griffiths writes,
if you're only interested in the field due to the essentially flat local patch of surface charge itself, the answer is...
Griffiths' Electrodynamics says that the electric field of a uniformly charged infinite plane, surface charge density sigma, is sigma/2e0. The field of an infinite sheet of charge is said to be sigma/e0, twice that of the plane.
What is the supposed difference between the sheet and the plane...
Homework Statement
I'm working my way through Classical Mechanics by J.R. Taylor. I'm stumped by the one-star ("easiest") Problem 5.11: "You are told that, at known positions x1 and x2, an oscillating mass m has speeds v1 and v2. What are the amplitude and angular frequency of the...
The answer is that this collision must be elastic. The least amount of energy that a stationary ground-state H atom can absorb is 10.2ev, to change the principal quantum number n from 1 (-13.6ev) to 2 (-3.4ev). Such a change would result in a residual K' = 12.0 - 10.2 = 1.8ev, which is less than...
TSny embarrasses me into working out the math. I find that my hand-waving was wrong. Inelastic collision is possible if at least half of the initial kinetic energy remains as kinetic energy after the collision. That is, up to half the initial K can change the atom's internal state if appropriate...
Homework Statement
Ground-state hydrogen atom with 12 ev kinetic energy collides head-on with another ground-state hydrogen atom at rest. Using principles of conservation of energy and momentum, show that an inelastic collision cannot occur. Therefore the collision must be elastic...