Answer:
[tex]4.87772175\times 10^{-15}\ C/m^2[/tex]
2 mm
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
[tex]\Delta l[/tex] = Distance charge traveled = 2 mm
v = Velocity of electron = 450 m/s
E = Electric field
[tex]m_e[/tex] = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
[tex]q_e[/tex] = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
As the energy of the system is conserved we have
[tex]q_eE\Delta l=\dfrac{1}{2}m_ev^2\\\Rightarrow E=\dfrac{1}{2}\dfrac{m_e}{q_e}\times \dfrac{v^2}{\Delta l}\\\Rightarrow E=\dfrac{1}{2}\dfrac{9.11\times 10^{-31}}{1.6\times 10^{-19}}\times \dfrac{450^2}{2\times 10^{-3}}\\\Rightarrow E=0.00028824609375\ N/C[/tex]
For an infinite non conducting sheet electric field is given by
[tex]E=\dfrac{\sigma}{2\epsilon}\\\Rightarrow \sigma=2E\epsilon\\\Rightarrow \sigma=2\times 0.00028824609375\times 8.85\times 10^{-12}\\\Rightarrow \sigma=5.1019558594\times 10^{-18}\ C/m^2[/tex]
The surface charge density is [tex]4.87772175\times 10^{-15}\ C/m^2[/tex]
The surface charge density is constant throughout so the electron should be fired from the distance of 2 mm