A pair of parallel plates at 4 cm apart is connected to a battery. When a proton is placed at a point 1 cm from the positive plate, the electric force on the proton is 1.6x10^-5 N. When the proton is placed at a point 1cm from the negative plate,
a) what is the electric force on the proton?
b) what is the p.d. between the final and initial positions of the proton?
take e=1.6x10^-19 C.
potential different
2011-09-26 7:06 am
回答 (1)
2011-09-26 8:33 pm
✔ 最佳答案
(a) Since the electric field strength in between the parallel plates is uniform, the force on the proton remains the same, i.e. 1.6x10^-5 N.(b) Since force on proton = charge of proton x electric field intensity
1.6x10^-5 = (1.6x10^-19).E
where E is the electric field intensity
i.e. E = 1.6x10^-5/1.6x10^-19 v/m = 1 x 10^14 v/m
Because the separation between the two positions of the proton is 2 cm (0.02 m)
hence, p.d. = (1x10^14) x 0.02 v = 2 x 10^12 v
收錄日期: 2021-04-20 11:33:58
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