Answer:
528398.4375 N/C opposite to the direction of the proton
[tex]3.56\times 10^{-8}\ s[/tex]
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 = [tex]1.8\times 10^{6}\ m/s[/tex]
s = Displacement = 3.2 cm
a = Acceleration
Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
Mass of electron = [tex]1.67\times 10^{-27}\ kg[/tex]
q = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]
[tex]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[/tex]
Electric field is given by
[tex]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[/tex]
The electric field is 528398.4375 N/C opposite to the direction of the proton
[tex]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[/tex]
The time taken is [tex]3.56\times 10^{-8}\ s[/tex]
[tex]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[/tex]
The electric field is 288.24609375 N/C in the same direction of the motion of the electron
