標題:
AL Physics MC
發問:
As follows: 圖片參考：http://hk.geocities.com/stevieg_1023/0RESS.gif
最佳解答:
At steady state, there will be no current through the capacitor C. So, current flows only through the resistors. Therefore, R1, R2 connected in series. R3 and R4 connected in series. And the two branches are in parallel. In series, p.d. across R1 = 6 / (3 + 6) X 6 = 4 V p.d. across R2 = 6 - 4 = 2 V On the other hand, p.d. across R3 = 1 / (1 + 2) X 6 = 2 V p.d. across R4 = 6 - 2 = 4 V Hence, B is at a higher potential than A, potential difference across C is = 4 - 2 = 2 V, with B at a higher potential. Therefore, negative charges accumulate on the plate of the capacitor connecting to A. And the charges, Q = CV = (20 X 10^-6)(-2) = -40 X 10^-6 C So, the answer is E.
其他解答:
AL Physics MC
發問:
As follows: 圖片參考：http://hk.geocities.com/stevieg_1023/0RESS.gif
最佳解答:
At steady state, there will be no current through the capacitor C. So, current flows only through the resistors. Therefore, R1, R2 connected in series. R3 and R4 connected in series. And the two branches are in parallel. In series, p.d. across R1 = 6 / (3 + 6) X 6 = 4 V p.d. across R2 = 6 - 4 = 2 V On the other hand, p.d. across R3 = 1 / (1 + 2) X 6 = 2 V p.d. across R4 = 6 - 2 = 4 V Hence, B is at a higher potential than A, potential difference across C is = 4 - 2 = 2 V, with B at a higher potential. Therefore, negative charges accumulate on the plate of the capacitor connecting to A. And the charges, Q = CV = (20 X 10^-6)(-2) = -40 X 10^-6 C So, the answer is E.
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