GREPhysics.NET: GR9677 #24

GR9677 #24

Problem

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Electromagnetism $\Rightarrow$ }Conductors

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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Comments

memyselmineni
2014-11-27 05:18:57

http://www.amazon.com/Physics-Mathematica-Jude-Ndubuisi-Onicha/dp/1499691920/ref=sr_1_fkmr1_1?s=books&ie=UTF8&qid=1401627044&sr=1-1-fkmr1&keywords=physics+mathematica++2nd+edition+by+onicha+jude

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dragore
2012-08-19 13:48:01

Y'all wish the whole test was that easy, I know.

OneWeekLeft
2016-10-23 23:58:51

lol so true

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Y'all wish the whole test was that easy, I know.

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