To the right is a diagram of two parallel charged plates. (pic: it is just two p

Question

To the right is a diagram of two parallel charged plates. (pic: it is just two parallel charged plates with the postive particles on the left and negative particles on the right) The separation of the plates is 1.2 cm. an electron leaves the negative plate at zero velocity and passes through a small hole in the positive plate. the electric field strength in the region between the plates is 2.2*10^4. a. determine the voltage connecting the plates. b. if there are 3 uF capacitance in the plates, what is the charge on the plates? c.determine the force the electron experiences while in between the plates. d. determine the electric potential energy the electron has when it starts. e. determine the veleocity the electron has as it passes through the positive plate.

Explanation / Answer

a) V = Ed = 2.2E4*1.2E-2=264 V

b) Q = CV = 3 E-6*264=0.000792 C

c) F = q E = 1.6E-19*2.2E4=3.52E-15 N

d) PE = q V = 1.6E-19*264=4.224E-17

e) 1/2 mv^2 = PE
0.5*9.11E-31*v^2=4.224E-17
v=9.63E6 m/s

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