Answer:
[tex]6.945326087\times 10^{-11}\ C[/tex]
5434.78260873 N/C
2963369.48874 m/s
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
V = Voltage = 25 V
Side = 3.8 cm
d = Separation = 4.6 mm
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
Charge is given by
[tex]Q=\dfrac{VA\epsilon_0}{d}\\\Rightarrow Q=\dfrac{25\times (3.8\times 10^{-2})^2\times 8.85\times 10^{-12}}{4.6\times 10^{-3}}\\\Rightarrow Q=6.945326087\times 10^{-11}\ C[/tex]
Charge on each plate is [tex]6.945326087\times 10^{-11}\ C[/tex]
Electric field is given by
[tex]E=\dfrac{Q}{A\epsilon_0}\\\Rightarrow E=\dfrac{6.945326087\times 10^{-11}}{(3.8\times 10^{-2})^2\times 8.85\times 10^{-12}}\\\Rightarrow E=5434.78260873\ N/C[/tex]
The electric field strength is 5434.78260873 N/C
Acceleration is given by
[tex]a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 5434.78260873}{9.11\times 10^{-31}}\\\Rightarrow a=9.5451725291\times 10^{14}\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.5451725291\times 10^{14}\times 4.6\times 10^{-3}+0^2}\\\Rightarrow v=2963369.48874\ m/s[/tex]
The velocity is 2963369.48874 m/s
