Respuesta :
Answer:
The velocity is [tex]2.94\times10^{6}\ m/s[/tex].
Explanation:
Given that,
Charge density of the plane [tex]\sigma=7.65\times10^{−4}\ C/m^2[/tex]
Distance = 1.05 mm
We need to calculate the electric field due to plane of charge
Using formula of electric field
[tex]E=\dfrac{\sigma}{2\epsilon}[/tex]
Put the value into the formula
[tex]E=\dfrac{7.65\times10^{−4}}{2\times8.85\times10^{-12}}[/tex]
[tex]E=4.322\times10^{7}\ N/C[/tex]
We need to calculate potential difference
Using formula of potential difference
[tex]V=E\times r[/tex]
Put the value into the formula
[tex]V=4.322\times10^{7}\times1.05\times10^{-3}[/tex]
[tex]V=4.5381\times10^{4}\ Volt[/tex]
We need to calculate the work requires to be done to reach the surface of the plane
Using formula of work done
[tex]W=qV[/tex]
Put the value into the formula
[tex]W = 1.6\times10^{-19}\times4.5381\times10^{4}[/tex]
[tex]W=7.26096\times10^{-15}\ J[/tex]
We need to calculate the velocity
Using work energy theorem
[tex]W=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=\dfrac{2W}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times7.26096\times10^{-15}}{1.67\times10^{-27}}}[/tex]
[tex]v=2.94\times10^{6}\ m/s[/tex]
Hence, The velocity is [tex]2.94\times10^{6}\ m/s[/tex].
