Solution 1 - Theoretical Question 1 / Solutions Page 1/12...

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Theoretical Question 1 / Solutions Page 1/12 Theoretical Question 1: Ping-Pong Resistor 1. Answers (a) 2 2 0 2 R 2 1 d V R F ε π = (b) d r 2 0 χ = (c) mgd V 2 th = (d) β α + = 2 s v V = m η 2 1 2 2 , () gd 2 1 2 2 + = (e) 2 3 2 1 1 md γ + = (f) mgd V c 2 2 1 1 + = , m g I c ) 1 )( 1 ( 1 2 2 2 + + = V/ V th I 0 c z 1 I c 2 ~ V I
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Theoretical Question 1 / Solutions Page 2/12 2. Solutions (a) [1.2 points] The charge Q induced by the external bias voltage V can be obtained by applying the Gauss law: = Q s d E r r 0 ε ( a 1 ) ) ( ) ( 2 0 2 0 R d V R E Q π = = , (a2) where Ed V = . The energy stored in the capacitor: d V R V d d V R V d V Q U V V 2 2 0 0 2 0 0 2 1 ) ( = = = . (a3) The force acting on the plate, when the bias voltage V is kept constant: 2 2 2 0 R 2 1 d V R d U F = + = . (a4) [An alternative solution:] Since the electric field ' E acting on one plate should be generated by the other plate and its magnitude is d V E E 2 2 1 ' = = , ( a 5 ) the force acting on the plate can be obtained by ' R QE F = . ( a 6 ) (b) [0.8 points] The charge q on the small disk can also be calculated by applying the Gauss law: = q s d E r r 0 . ( b 1 ) Since one side of the small disk is in contact with the plate, V V d r r E q χ = = = 2 0 2 0 ) ( . (b2)
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Theoretical Question 1 / Solutions Page 3/12 Alternatively, one may use the area ratio for Q R r q = 2 2 π . d r 2 0 ε χ = . ( b 3 ) (c) [0.5 points] The net force, net F , acting on the small disk should be a sum of the gravitational and electrostatic forces: e g net F F F + = . ( c 1 ) The gravitational force: mg F = g . The electrostatic force can be derived from the result of (a) above: 2 2 2 2 0 e 2 2 1 V d V d r F = = . (c2) In order for the disk to be lifted, one requires : 0 net > F 0 2 2 > mg V d . (c3) mgd V 2 th = . (c4) (d) [2.3 points] Let s v be the steady velocity of the small disk just after its collision with the bottom plate. Then the steady-state kinetic energy s K of the disk just above the bottom plate is given by 2 s s v 2 1 m K = . ( d 1 ) For each round trip, the disk gains electrostatic energy by qV U 2 = . ( d 2 ) For each inelastic collision, the disk lose its kinetic energy by after 2 before 2 after before loss 1 1 ) 1 ( K K K K K = = = η . (d3) Since s K is the energy after the collision at the bottom plate and ) ( s mgd qV K + is
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Theoretical Question 1 / Solutions Page 4/12 the energy before the collision at the top plate, the total energy loss during the round trip can be written in terms of s K : ) )( 1 ( 1 1 s 2 s 2 tot mgd qV K K K + + = η . (d4) In its steady state, U should be compensated by tot K . ) )( 1 ( 1 1 2 s 2 s 2 mgd qV K K qV + + = . (d5) Rearranging Eq. (d5), we have [] . v 2 1 1 1 ) 1 ( ) 1 ( 1 2 s 2 2 2 2 2 2 4 2 s m mgd qV mgd qV K = + + = + + = (d6) Therefore, () gd m V 2 1 2 1 v 2 2 2 2 2 s + +
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This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.

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Solution 1 - Theoretical Question 1 / Solutions Page 1/12...

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