Question

A proton falls down through a distance of 2
cm in a uniform electric field of magnitude 3.34
x 103 NC-1.
I
Determine (i) the acceleration of the electron
(ii) the time taken by the proton to fall through
the distance of 2 cm,​

Answer 1

Answer:

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Answer 2

Answer:

Your question is “A proton falls down through a distance of uniform electric field of 3.34×10×10×10 N/C. What is the acceleration of electron the time taken by the proton to fall through the distance of 2cm and the direction of electric field?”

Did you mean to say “A proton accelerates through a distance of 2 cm in a uniform electric field, starting from rest …….. What is the acceleration of the proton, the time taken to travel 2 cm, …..etc..”?

We assume that you meant to say something similar, perhaps with a better turn of phrase.

A proton has a charge q = +1.6 E-19 C so the force F on the proton in an electric field E would be F = q E which is in the same direction as E. Since the magnitude of the electric field is known, E = 3.34 E+3 N/C, the magnitude of the force F = [+1.6 E-19 C ] * [3.34 E+3 N/C] which is F = 5.344 E-16 N ……..(1)

We also know that force = mass x acceleration, that is

F = m(p) * a ……….. (2)

and we know m(p) = 1.67 E-27 kg

So acceleration = (5.344/1.67) E+11 m sec^-2

Using the laws of motion we can find any other parameters as desired.

The direction of the field is anyone's choice.