A uniform electric field exists in a region between two oppositely-charged plates. An electron is released from rest at the surface of the negatively-charged plate and strikes the surface of the opposite plate, d = 2 cm away, 20 ns later. a) What is the speed of the electron as it strikes the second plate? b) What is the magnitude of the electric field? c) If instead, a proton is released from rest at the positively charged plate, how long would it take to get to the opposite plate?

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 2\times 10^{-2}=0\times t+\frac{1}{2}\times a\times (20\times 10^{-9})^2\\\Rightarrow a=\frac{2\times 10^{-2}\times 2}{(20\times 10^{-9})^2}\\\Rightarrow a=1\times 10^{14}\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow v=0+1\times 10^{14}\times 20\times 10^{-9}\\\Rightarrow v=2000000\ m/s[/tex]

The speed of the electron is 2000000 m/s

Electric field is given by

[tex]E=\dfrac{ma}{q}\\\Rightarrow E=\dfrac{9.11\times 10^{-31}\times 1\times 10^{14}}{1.6\times 10^{-19}}\\\Rightarrow E=569.375\ N/C[/tex]

The electric field is 569.375 N/C

m = Mass of proton = [tex]1.67\times 10^{-27}\ kg[/tex]

[tex]E=\dfrac{ma}{q}\\\Rightarrow a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{569.375\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=5.4550898204\times 10^{10}\ m/s^2[/tex]

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 0.02=0t+\frac{1}{2}\times 5.4550898204\times 10^{10}\times t^2\\\Rightarrow t=\sqrt{\frac{0.02\times 2}{5.4550898204\times 10^{10}}}\\\Rightarrow t=8.5630610979\times 10^{-7}\ s[/tex]

The time taken is [tex]8.5630610979\times 10^{-7}\ s[/tex]