Question

A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0. The proton travels 7.00 cm as it comes to rest. Determine (a) the acceleration of the proton, (b) its initial speed, and (c) the time interval over which the proton comes to rest.

Answer 1

(a). The magnitude of the acceleration of the proton is
`5.74*10^(13)\ m/s^2`

(b). The initial peed of the protion is
`2.83*10^(6)\ m/s`

(c). The time is
`0.493*10^(-7)\ sec`.

Step-by-step explanation:

Given that,

Electric field
`E=-6.00*10^(5)i\ N/C`

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

`F=F_(e)`

`ma=qE`

`a=(Eq)/(m)`

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

`a=(-6.00*10^(5)*1.6*10^(-19))/(1.67*10^(-27))`

`a=-5.74*10^(13)\ m/s62`

The magnitude of the acceleration of the proton is
`5.74*10^(13)\ m/s^2`

(b). We need to calculate the initial peed

Using equation of motion

`v^2-u^2=2as`

Where, s = distance

Put the value into the formula

`0-u^2=2*(-5.74*10^(13))*7.00*10^(-2)`

`u=\sqrt{2*5.74*10^(13)*7.00*10^(-2)}`

`u=2834783.94`

`u=2.83*10^(6)\ m/s`

The initial peed of the protion is
`2.83*10^(6)\ m/s`

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

`t=(u)/(a)`

Where, u = initial velocity

a = acceleration

Put the value into the formula

`t=(2.83*10^(6))/(5.74*10^(13))`

`t=0.493*10^(-7)\ sec`

Hence, (a). The magnitude of the acceleration of the proton is
`5.74*10^(13)\ m/s^2`

(b). The initial peed of the protion is
`2.83*10^(6)\ m/s`

(c). The time is
`0.493*10^(-7)\ sec`.