Use the figure at the right to answer questions 11 and 12 Assume that the capaci

Question

Use the figure at the right to answer questions 11 and 12 Assume that the capacitor is completely uncharged before switch s is closed Referring to the circuit, " a long time after the switch S is closed" means that more than 30 minutes pass after the switch is closed. Immediately after the switch S is closed, the capacitor behave as if there is a "wire connection(i.e., a short circuit") in the branch between points alpha and beta. Immediately after the switch S is closed the capacitor behaves as if the lower end of R_2 is left unconnected from the circuit, A long time after the switch is closed the current in R_2 is at the maximum value. A long time after the switch is closed the current in R_2 is exactly the same as the current in R_1. Switch S has been closed for a very long time. It is then opened and C discharges. How long after S is opened before C can be considered to e completely discharged?

Explanation / Answer

q = 3e=3x1.6x10-19 C, v =8.x107 m/s, B =150 mT, theta =15.8 degrees

F=qvBsin(theta)

F =(3x1.6x10-19)x(8.x107)x(0.15)xsin(15.8)

F =1.61x10-12 N

F =1.61 N

F=qvxB

v is along x direction

B is along y direction

F = ixj = k (out of page)

F=1.61 pN Outof page

Option (a) is correct

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