An electron is projected with an initial speed v 0 = 2.00×10 6 m/s into the unif

Question

An electron is projected with an initial speed v0 = 2.00×106 m/s into the uniform field between the parallel plates in the figure. 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.

Part A

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Part B

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?

Part C

What would be the direction of proton's displacement?

displacement is downward

Part D

Compare the paths traveled by the electron and the proton and explain the differences.

displacement is downward

Explanation / Answer

A-) Initial horizontal velocity of electron = 2 x 10^6 m/s

Initial vertical velocity = 0 m/s

As electric field is along yaxis, there wil no force acting on electron along x-axis.

time taken by electron to cross the plate ( horizontal distance)= (0.02)/( 2 x 10^6)= 0.01 x 10^-6 sec apprx

acceleration of electron can be calcualted using kinematic equation:

h = 1/2at^2

h = 0.5cm ( half of 1cm) = 0.005 m

0.005 = 0.5 ( a)( 0.01 x 10^-6 )^2

a = 100 x 10^12 m/s ^2

F= qE = ma

E ( electric field) = ma/q

mass of electron = 9.11 x 10^-31 kg, q = charge on electron =-1.6 x 10^-19 C

E = 9.11 x 10^-31 x 100 x 10^12/ 1.6 x 10^-19= 5.693 x10^2 N/C apprx

b) Acceleratiion for proton = qE/ m = 1.6 x 10^-19 x 5.693x10^2 / ( 1.6726219 × 10-27 ) where mass of proton = 1.6726219 × 10-27 kg

a= 5.446 x 10 ^ 10 m/s ^2

h eight gained by proton = 0.5 ( 5.446 x 10 ^ 10 )( 0.01 x 10^-6 )^2

Proton will not hit the plate.

c) verrtical displacement of proton = 0.000272 cm apprx

d) Direction = towrads the bottom plate

e) electron will be deflected towards the upper plate as it has negative charge , while proton will be deflected towards the bottom plate.

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