An electron, moving at a speed of 7.64*10^6m/s, enters the region between two pa

Question

An electron, moving at a speed of 7.64*10^6m/s, enters the region between two parallel plates seperated by a distance of 5.22cm. As it enters, it is moving in a direction parallel to the plates themselves, and is 3.55cm from the positive plate. Between the plates is a potential difference of 964V. The mass of the electron is 9.11*10^-31kg.

a) find is velocity in the direction perpendicular to the plane of the plates at the timewhen it strikes the positive plate.

b)During that time, how far will the electron have traveled in the direction parallel to the plates?

Explanation / Answer

F = qE
Now, V = E/d
E = 964*0.0522 = 18467.43295 N/C
F = qE = ma
a = 3.247*10^15 m/s/s
0.0355 = 0.5*a*t^2
t = 4.6761*10^-9 s

a) v = at = 1.5183395*10^7 m/s
b) s = vt = 0.035725 m

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