Question

Calculate the speed of a proton that has moved in a uniform electric field of 180.0 N/C from rest over a distance of 12.5-cm. Assume it began at rest.

Answer 1

the velocity of the proton is 65574.38 m/s

Step-by-step explanation:

given,

uniform electric field = 180 N/C

Distance = 12.5 cm = 0.125 m

charge of proton = 1.6 × 10⁻¹⁹ C

force = E × q

=180 × 1.6 × 10⁻¹⁹

F= 2.88 × 10⁻¹⁷ N

mass of proton = 1.673 × 10⁻²⁷ kg

acceleration =
`(force)/(mass)`

=
`(2.88 * 10^(-17))/(1.673* 10^(-27))`

=1.72 × 10¹⁰ m/s²

velocity =
`\sqrt{2* 0.125 * 1.72 * 10^(10)}`

=65574.38 m/s

hence , the velocity of the proton is 65574.38 m/s

Answer 2

Given:

Electric field = 180 N/C

`Force\ on\ proton = 1.6*10^(-19) C`

`Force\ on\ proton = 180*1.6*10^(-19) =288*10^(-19) N`

`Mass\ of\ proton = 1.673*10^(-27) kg`

`Acceleration of proton = (force)/(mass)`

`Acceleration\ of\ proton = (288*10^(-19))/(1.673*10^(-27)) =172*108 m/s^(2)`

Let the speed of proton be "x"

x =
`√(Acceleration)`

`x = √((2*172*108*0.125))=65602.2 m/s`