An electron is projected with an initial speed v_0 = 1.10 times 10^8 m/s into th

Question

An electron is projected with an initial speed v_0 = 1.10 times 10^8 m/s into the uniform field between the parallel plates in the figure (Figure 1). Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. If the proton would not hit one of the plates, what would be the magnitude of its vertical displacement as it exits the region between the plates? What would be the direction of proton's displacement? displacement is upward displacement is downward Compare the paths traveled by the electron and the proton and explain the differences.

Explanation / Answer

Time taken to cross the plates , t = 0.02/(1.10×106) =18 ns

Using equation of motion :

S= ut+(1/2)at2

Given that electron enters midway between the plates , and it just misses the upper plate as it emerges. Hence vertical displacement of the electron is 0.005 m.(0.5cm)

Using equation of motion :

S = ut+(1/2)at2

0.005= 0+(1/2)a (18×10-9)2

a = 0.3086× 1014m/s2

i.e. E = ma/e = (9.1×10-31)(0.3086×1014)/(1.6×10-19)

E = 175.51 V/m

Part:C

The foce acting on proton due to electric field : F = eE

i.e. eE= ma ,

Acceleration , a =eE/m

= (1.6×10-19)(175.51)/(1.67×10-27)

= 1.68× 108 m/s2

The proton takes same time as electron to cross the plates.

Vertical displacement

S= ut+(1/2)at2

= 0+(1/2)(1.68×108)(18×10-9)2

= 2.72 ×10-8 m

Part D :

The vertical displacement is in downward direction

PartE:

the path will be different as for proton vertical displacement is in downward direction while for electron it is upward.

Write down situation given in picture in your own words.

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