....o Zain KSA F 4:48 PM bbappsrv.pmu.edu.sa Two completely discharged capacitor
ID: 1489943 • Letter: #
Question
....o Zain KSA F 4:48 PM bbappsrv.pmu.edu.sa Two completely discharged capacitors that are NOT identical are connected in series. This series combination is then connected across the terminal of a battery to charge the two capacitors. When the current stops, the quantity that is the same for both capacitors is: a) potential difference. b) stored energy. c energy density. d) electric field. e) charge on the positive plates. A point charge q is placed at the center of a spherical Gaussian surface. The electric flux through the surface will change if a) the radius of the surface is doubled. b) the shape of the surface is changed to a cube with the same volume as the original sphere. c) the point charge is moved off center (but still inside the sphere) d) the point charge is moved to a location just outside the sphere. e) if a second point charge is placed just outside the sphere. A 30-turn solenoid is 0.120 m long. The solenoid has a resistance of13.5 mm and is connected to a 20.0 V battery. The magnitude of the magnetic field at the center of the solenoid is a 400 mT b) 465 mT c) 556 m T e) 167 mT d) 89.0 mT When a positive charge moves in the direction of an electric field, a) the charge increases b) the field does work on the charge. c) the charge decreases. d) the charge gains potential energy. e) the charge loses kinetic energy. Question xx and xx refer to the RC circuit shown. The time constant for this circuit is a) 0.5 s b) 1.0 c) 1.5 s d) 2.0 s e) 3.0 s After a long time, the switch is moved from position a to position b. 3.0 s later, the energy stored in the capacitor is a) 4 mJ b) 8 mJ c) 11 mJ d 29 mJ e) 80 mJExplanation / Answer
e) charge on the positive plates
d) the point charge is moved to a location just outside the sphere.
b) B = u0NI/L = 4pi x 10-7 x 30 x 20/(0.0135 x 0.12) = 0.4654 T = 465.4 mT
b) the field does work on the charge
c) time constant = RC = 50 x 103 x 30 x 10-6 = 1.5 s
a) q = CV0e-t/1.5 = 30 x 10-6 x 120 x e-2 = 0.4872 x 10-3
energy = q2/2C = 3.96 mJ
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