HW3 solutions
are out for your studying pleasure.
UPDATE: I corrected a dumb mistake in 3a - I forgot to multiply by 4 when I collected all of the q5 terms, so the answer should be 4-15*sqrt(2) times the energy of a single pair, or about -17.21 as pointed out in the comments. The link above has the corrected version.
UPDATE: Small mistake in 1b as well - the initial formula for acceleration is OK, but the final expression where I've plugged in the result for d was incorrect, as was the numerical answer. Fixed now.
UPDATE: I corrected a dumb mistake in 3a - I forgot to multiply by 4 when I collected all of the q5 terms, so the answer should be 4-15*sqrt(2) times the energy of a single pair, or about -17.21 as pointed out in the comments. The link above has the corrected version.
UPDATE: Small mistake in 1b as well - the initial formula for acceleration is OK, but the final expression where I've plugged in the result for d was incorrect, as was the numerical answer. Fixed now.
6 comments:
THANKS!!
Can you explain the answer for 3a. When you worked it out in class you got different numbers for the charges that involved the -4q.
In 3a isn't there supposed to be 4 of the (Keq^2)/(a)(sqrt2/2) so would the answer be -17.21 instead of -5.9?
Ah, my bad. I dropped a factor 4, the last one should be -17.21. Will repost the file with corrections. Should not do these things so late at night ...
I don't understand why #1 part b is worked out the way that it is. I can't get get the right number even working through it exactly. Also, why is acceleration in newtons and not m/s^2? Could you please explain?
You're right - see update to top level post. The initial expression is fine, but I made a typo in plugging in the result for d. This made the numerical answer incorrect as well. The units should be m/s^2 as you point out, also fixed.
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