Exam2_2010Spr_Solutions

Exam2_2010Spr_Solutions - MASSACHUSETTS INSTITUTE OF...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2010 8.02 Exam Two Spring 2010 Solutions Question 1: (5 points) Circle the correct answer. Consider a simple parallel-plate capacitor whose plates are given equal and opposite charges and are separated by a distance d. The capacitor is connected to a battery. Suppose the plates are pushed together until they are separated by a distance D = d/2. How does the final electrostatic energy stored in the capacitor compare to the initial energy? a) Final is half the initial. b) Final is one fourth the initial. c) Final is twice than initial. d) Final is four times the initial. e) They are the same. Answer c. Because the capacitor is connected to the battery, the potential difference across the plates is constant. Therefore the ratio of the final stored energy to the initial stored energy is proportional ratio of the final capacitance to the initial capacitance, 22 11 /// f if C i C f i UU CV CV CC Δ = . For a parallel plate capacitor, the capacitance is inversely proportional to the distance separating the plates, 0 0 (/ ) A QQ A C VE d d d ε σ σε === = Δ . Therefore the ratio of the capacitance is ( / 2 ) 2 fi CC dd dd == = . So /2 = .
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Question 2: (5 points) Circle the correct answer. A conducting wire is attached to an initially charged spherical conducting shell of radius 2 a . The other end of the wire is attached to the outer surface of a neutral conducting spherical shell of radius a that is located a very large distance away (at infinity). When electrostatic equilibrium is reached, the charge on the shell of radius 2 a is equal to a) one fourth the charge on the shell of radius a . b) half the charge on the shell of radius a . c) twice the charge on the shell of radius a . d) four time the charge on the shell of radius a . e) None of the above. Answer c. When electrostatic equilibrium is reached, the two shells form one conducting surface and hence the potential on that surface is constant. Because the two shells are very far apart, the potential of each shell with respect to infinity can be calculated separated. Because the electric field outside each charged shell is identical to the electric field of a point-like object with the same charge located at the center of the shell. The potential on each shell with respect to infinity is just 0 /4 Qr πε . The potential difference between the two shells is zero, or 20 0 2 0 aa Qa Q a π επ ε = , or 2 2 QQ = . Therefore the charge on the shell of radius 2 a is equal to twice the charge on the shell of radius a .
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Question 3: (5 points) Circle the correct answer. What is the correct order for the total power dissipated in the following circuits, from least to greatest? Assume all bulbs and all batteries are identical. Ignore any internal resistance of the batteries.
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This note was uploaded on 02/27/2011 for the course PHYSICS 8 taught by Professor Dourmashkin during the Fall '10 term at MIT.

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Exam2_2010Spr_Solutions - MASSACHUSETTS INSTITUTE OF...

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