Question

A thin conducting square plate 1.0 m on the side is given a charge of-2.0 x 10-6 c. A proton is placed 1.0 en above the center of the plate, what is the acceleration of the proton? (Enter the magnitude in m/s.) magnitude direction Select the plate m/s2

Answer 1

Answer:

Acceleration, $a=1.08\times 10^{13}\ m/s^2$

Explanation:

It is given that,

Side of the square plate, l = 1 m

Charge on the square plate, $Q=-2\times 10^{-6}\ C$

Position of a proton, x = 1 cm

The electric field due to a parallel plate is given by :

$E=\dfrac{Q}{2A\epsilon_o}$

Electric force is given by :

F = q E

$F=\dfrac{Qe}{2A\epsilon_o}$

e is the charge on electron

The acceleration of the proton can be calculated as :

$a=\dfrac{F}{m}$

m is the mass of proton

$a=\dfrac{Qe}{2A\epsilon_o m}$

$a=\dfrac{2\times 10^{-6}\times 1.6\times 10^{-19}}{2(1)^2\times 8.85\times 10^{-12}\times {1.67\times 10^{-27}}}$

$a=1.08\times 10^{13}\ m/s^2$

So, the acceleration of the proton is $1.08\times 10^{13}\ m/s^2$. Hence, this is the required solution.