Answer :
Explanation:
Formula to calculate the electric field of the sheet is as follows.
E = [tex]\frac{\sigma}{2 \epsilon_{o}}[/tex]
And, expression for magnitude of force exerted on the electron is as follows.
F = Eq
So, work done by the force on electron is as follows.
W = Fs
where, s = distance of electron from its initial position
= (0.570 - 0.06) m
= 0.51 m
First, we will calculate the electric field as follows.
E = [tex]\frac{\sigma}{2 \epsilon_{o}}[/tex]
= [tex]\frac{4.60 \times 10^{-12}C/m^{2}}{2 \times 8.854 \times 10^{-12}C^{2}/N m^{2}}[/tex]
= 0.259 N/C
Now, force will be calculated as follows.
F = Eq
= [tex]0.259 N/C \times 1.6 \times 10^{-19} C[/tex]
= [tex]0.415 \times 10^{-19} N[/tex]
Now, work done will be as follows.
W = Fs
= [tex]0.415 \times 10^{-19} N \times 0.51 m[/tex]
= [tex]2.12 \times 10^{-20} J[/tex]
Thus, we can conclude that work done on the electron by the electric field of the sheet is [tex]2.12 \times 10^{-20} J[/tex].
Answer:
The work done is [tex]2.12\times10^{-20}\ J[/tex] and 0.1325 eV.
Explanation:
Given that,
Distance = 0.570 m
Surface charge density [tex]\sigma=4.60\times10^{-12}\ C/m^{2}[/tex]
Using formula of electric field
[tex]E = \dfrac{\sigma}{2\epsilon_{0}}[/tex]
Using formula of electrostatic force ,
[tex]F = eE[/tex]
[tex]F=e\times(\dfrac{\sigma}{2\epsilon_{0}})[/tex]
(A). We need to calculate the work done
Using formula of work done
[tex]W=Fd[/tex]
[tex]W=e\times(\dfrac{\sigma}{2\epsilon_{0}})d[/tex]
Put the value into the formula
[tex]W=1.6\times10^{-19}\times(\dfrac{4.60\times10^{-12}}{2\times8.85\times10^{-12}})\times(0.570-6.00\times10^{-2})[/tex]
[tex]W=2.12\times10^{-20}\ J[/tex]
[tex]W=\dfrac{2.12\times10^{-20}}{1.6\times10^{-19}}\ ev[/tex]
[tex]W=0.1325\ eV[/tex]
Hence, The work done is [tex]2.12\times10^{-20}\ J[/tex] and 0.1325 eV.