Question

A proton is traveling horizontally to the right at 4.20×10^6m/s.Part A:Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50cm.Part B: counterclockwise from the left directionPart C:How much time does it take the proton to stop after entering the field?Part D:What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?

Answer 1

2630250 N/C, horizontally left

0

`1.67* 10^(-8)\ s`

1434.825 N/C, horizontally left

Step-by-step explanation:

m = Mass of particle

u = Initial velocity =
`4.2* 10^6\ m/s`

v = Final velocity = 0

t = Time taken

s = Displacement = 3.5 cm

q = Charge of particle =
`1.6* 10^(-19)\ C`

Force is given by

`F=qE`

Acceleration is given by

`a=(F)/(m)\\\Rightarrow a=(qE)/(m)\\\Rightarrow a=(1.6* 10^(-19)* E)/(1.67* 10^(-27))\\\Rightarrow a=95808383.23353E`

`v^2-u^2=2as\\\Rightarrow v^2-u^2=2* 95808383.23353Es\\\Rightarrow E=(0^2-(4.2* 10^6)^2)/(2* 95808383.23353* 0.035)\\\Rightarrow E=-2630250\ N/C`

Magnitude of electric field is 2630250 N/C

Direction is horizontally to the left

The angle counterclockwise from left is zero.

`a=(F)/(m)\\\Rightarrow a=(qE)/(m)\\\Rightarrow a=(1.6* 10^(-19)* -2630250)/(1.67* 10^(-27))\\\Rightarrow a=-2.52* 10^(14)\ m/s^2`

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(0-4.2* 10^6)/(-2.52* 10^(14))\\\Rightarrow t=1.67* 10^(-8)\ s`

The time taken is
`1.67* 10^(-8)\ s`

Acceleration is given by

`a=(F)/(m)\\\Rightarrow a=(qE)/(m)\\\Rightarrow a=(1.6* 10^(-19)* E)/(9.11* 10^(-31))\\\Rightarrow a=175631174533.4797E`

`v^2-u^2=2as\\\Rightarrow v^2-u^2=2* 175631174533.4797Es\\\Rightarrow E=(0^2-(4.2* 10^6)^2)/(2* 175631174533.4797* 0.035)\\\Rightarrow E=-1434.825\ N/C`

Magnitude of electric field is 1434.825 N/C

Direction is horizontally to the left