Question

How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?

Answer 1

Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.

An electron is released from rest at a distance of 0 m from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² . How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?

Answer:

Work=1.06×10⁻²¹J

Step-by-step explanation:

Given Data

Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²

Charge density σ=3.90×10⁻¹² C/m²

The electron moves a distance d=3.00×10⁻²m

Electron charge e=-1.6×10⁻¹⁹C

To find

Work done

Solution

The electric field due is sheet is given as

E=σ/2ε₀

`E=(3.90*10^(-12)C/m^(2) )/(2(8.85*10^(-12)C^(2) /N.m^(2) ))\\ E=0.22V/m`

Now we need to find force on electron

`F=eE\\F=(1.6*10^(-19)C )(0.22V/m)\\F=3.525*10^(-20)N`

Now for Work done on the electron

`W=F*d\\W=(3.525*10^(-20) N)(3.00*10^(-2)m)\\W=1.06*10^(-21)J`