Answer:
Acceleration, [tex]a=1.08\times 10^{13}\ m/s^2[/tex]
Explanation:
It is given that,
Side of the square plate, l = 1 m
Charge on the square plate, [tex]Q=-2\times 10^{-6}\ C[/tex]
Position of a proton, x = 1 cm
The electric field due to a parallel plate is given by :
[tex]E=\dfrac{Q}{2A\epsilon_o}[/tex]
Electric force is given by :
F = q E
[tex]F=\dfrac{Qe}{2A\epsilon_o}[/tex]
e is the charge on electron
The acceleration of the proton can be calculated as :
[tex]a=\dfrac{F}{m}[/tex]
m is the mass of proton
[tex]a=\dfrac{Qe}{2A\epsilon_o m}[/tex]
[tex]a=\dfrac{2\times 10^{-6}\times 1.6\times 10^{-19}}{2(1)^2\times 8.85\times 10^{-12}\times {1.67\times 10^{-27}}}[/tex]
[tex]a=1.08\times 10^{13}\ m/s^2[/tex]
So, the acceleration of the proton is [tex]1.08\times 10^{13}\ m/s^2[/tex]. Hence, this is the required solution.
