A proton is fired directly into a uniform electric field that is able to bring t

Question

A proton is fired directly into a uniform electric field that is able to bring the charge momentarily to rest. (a) What must be the orientation of the electric field relative to the velocity of the proton? (b) What type of charge configuration would generate a uniform electric field? (c) If the proton is initially traveling at 5% the speed of light, what field strength will stop the charge within 1.00 cm of entering the field region? (d) How much work does the electric field do in stopping the proton? mostly need help w c is q 1.6E-19?

Explanation / Answer

a) The direction of elctric field must be opposite to the direction of motion of proton.

b) when positive and negative charges are uniformly distributed on plates that are placed closely, the elctric field between the plates will be uniform.

c)

use Work-energy theorem,

Workdone by electruc force = change in kinetic energy

F*d*cos(180) = (1/2)*m*(vf^2 -vi^2)

-F*d = (1/2)*m*(0^2 - v^2)

-q*E*d = -(1/2)*m*v^2

==> E = (1/2)*m*v^2/(q*d)

= (1/2)*1.67*10^-27*(0.05*3*10^8)^2/(1.6*10^-19*0.01)

= 1.17*10^8 N/c

d) Workdone = change in kinetic energy

= -(1/2)*m*v^2

= -(1/2)*1.67*10^-27*(0.05*3*10^8)^2

= 1.88*10^-13 J

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