Question

A proton is traveling horizontally to the right at 1.8 × 106 m/s. (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 cm? (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the condition of part (a)?

Answer 1

Answer:

528398.4375 N/C opposite to the direction of the proton

$3.56\times 10^{-8}\ s$

288.24609375 N/C in the same direction of the motion of the electron

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity = $1.8\times 10^{6}\ m/s$

s = Displacement = 3.2 cm

a = Acceleration

Mass of electron = $9.11\times 10^{-31}\ kg$

Mass of electron = $1.67\times 10^{-27}\ kg$

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

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(1.8\times 10^6)^2}{2\times 0.032}\\\Rightarrow a=-5.0625\times 10^{13}\ m/s^2$

Electric field is given by

$E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{1.67\times 10^{-27}\times -5.0625\times 10^{13}}{1.6\times 10^{-19}}\\\Rightarrow E=−528398.4375\ N/C$

The electric field is 528398.4375 N/C opposite to the direction of the proton

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-1.8\times 10^6}{-5.0625\times 10^{13}}\\\Rightarrow t=3.56\times 10^{-8}\ s$

The time taken is $3.56\times 10^{-8}\ s$

$E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{9.11\times 10^{-31}\times -5.0625\times 10^{13}}{-1.6\times 10^{-19}}\\\Rightarrow E=288.24609375\ N/C$

The electric field is 288.24609375 N/C in the same direction of the motion of the electron