t2fall02sol

t2fall02sol - PHYSICS 2306 TEST#2 1 November 2002 5:00 PM...

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1. An ideal parallel-plate capacitor, with plate separation of 4 mm, has its plates connected across a 100 Volt battery. If the plates are circular with radius 4 cm, then the capacitance, when air is in the gap between the plates, is about: (1) 1 pF (2) 10 pF C = ε o A/d = 8.85 10 -12 ⋅π (0.04) 2 /[4 10 -3 ] = 11.1 10 -12 10 pF (3) 100 pF (4) 1 nF (5) 10 nF (6) 100 nF 2. For the situation described in problem #1, if the battery remains connected, and a dielectric material with dielectric constant 2.5 is inserted to fill the space between the plates, then the electric field in the dielectric will be: (1) 100 V/m (2) 250 V/m (3) 1000 V/m (4) 2500 V/m (5) 10000 V/m Since the battery remains connected the potential change is still 100 V; hence (6) 25000 V/m in the dielectric E = V/d = 100/4 10 -3 = 25 10 +3 = 25000 V/m 3. Consider the system of capacitors as shown in the figure below. The equivalent capacitance is: (1) 3 µ F 3 µ F 6 µ F (2) 4 µ F a 12 µ F b (3) 6 µ F (4) 8 µ F 4 µ F (5) 12 µ F 1/C 36 = 1/3 + 1/6 = 3/6; C 36 = 2 µ F . C 364 = C 36 + C 4 = 2 + 4 = 6 µ F (6) 18 µ F 1/C eq = 1/C 364 + 1/C 12 = 1/6 + 1/12 = 3/12; C eq = 4 µ F 4. For the situation described in problem #3, when a potential difference is applied between a and b, the charge on the 4 µ F capacitor is 32 µ C. Then the charge on the 3 µ F capacitor is: (1) 16 µ C V 4 = Q 4 /C 4 = 32/4 = 8 V; then V 36 = 8 V, so Q
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This note was uploaded on 03/19/2008 for the course PHYS 2306 taught by Professor Ykim during the Fall '06 term at Virginia Tech.

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t2fall02sol - PHYSICS 2306 TEST#2 1 November 2002 5:00 PM...

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