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

Answer:

2630250 N/C, horizontally left

0

$1.67\times 10^{-8}\ s$

1434.825 N/C, horizontally left

Explanation:

m = Mass of particle

u = Initial velocity = $4.2\times 10^6\ m/s$

v = Final velocity = 0

t = Time taken

s = Displacement = 3.5 cm

q = Charge of particle = $1.6\times 10^{-19}\ C$

Force is given by

$F=qE$

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{1.67\times 10^{-27}}\\\Rightarrow a=95808383.23353E$

$v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 95808383.23353Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 95808383.23353\times 0.035}\\\Rightarrow E=-2630250\ N/C$

Magnitude of electric field is 2630250 N/C

Direction is horizontally to the left

The angle counterclockwise from left is zero.

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times -2630250}{1.67\times 10^{-27}}\\\Rightarrow a=-2.52\times 10^{14}\ m/s^2$

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-4.2\times 10^6}{-2.52\times 10^{14}}\\\Rightarrow t=1.67\times 10^{-8}\ s$

The time taken is $1.67\times 10^{-8}\ s$

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{9.11\times 10^{-31}}\\\Rightarrow a=175631174533.4797E$

$v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 175631174533.4797Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 175631174533.4797\times 0.035}\\\Rightarrow E=-1434.825\ N/C$

Magnitude of electric field is 1434.825 N/C

Direction is horizontally to the left