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  GR9677 #24
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This problem involves applying Coulomb's Law F\propto q_A q_B/r_{AB}^2 to conductors. The charge travels from conductor to conductor and equilibriates instantaneously due to the requirement that two touching conductors must be at an equipotential. This means that if conductors 1 and 2 touch then their potentials are related by V_1=V_2. Because the problem involves spherical conductors, the potential has the form V\propto q_A q_B/r_{AB}.

The initial force between the two conductors is F, where q_A=q_B=Q.

After C is touched to A, the charge becomes q_A=Q/2=q_C, since each conductor shares the same charge out of a total of Q (to wit: each has half of the total charge).

When C is touched to B, the charge becomes q_C=3/4Q=q_B, since each conductor shares the same charge out of a total of Q+Q/2 (to wit: \frac{1}{2}3/2 Q = 3/4 Q for each conductor).

When C is removed, one calculates the force from Coulomb's law and the final charges on A and B determined above to be, F = 3/8 Q^2/r_{AB}^2=3/8 F, as in choice (D).

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2014-11-27 05:18:57
2012-08-19 13:48:01
Y'all wish the whole test was that easy, I know.
2016-10-23 23:58:51
lol so true

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