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

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

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

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

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

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