Question

An electron is released from rest at a distance of 0.570 m from a large insulating sheet of charge that has uniform surface charge density 4.60×10−12 C/m2 . Part A 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 6.00×10−2 m from the sheet?

Answer 1

Answer:

The work done is $2.12\times10^{-20}\ J$ and 0.1325 eV.

Explanation:

Given that,

Distance = 0.570 m

Surface charge density $\sigma=4.60\times10^{-12}\ C/m^{2}$

Using formula of electric field

$E = \dfrac{\sigma}{2\epsilon_{0}}$

Using formula of electrostatic force ,

$F = eE$

$F=e\times(\dfrac{\sigma}{2\epsilon_{0}})$

(A). We need to calculate the work done

Using formula of work done

$W=Fd$

$W=e\times(\dfrac{\sigma}{2\epsilon_{0}})d$

Put the value into the formula

$W=1.6\times10^{-19}\times(\dfrac{4.60\times10^{-12}}{2\times8.85\times10^{-12}})\times(0.570-6.00\times10^{-2})$

$W=2.12\times10^{-20}\ J$

$W=\dfrac{2.12\times10^{-20}}{1.6\times10^{-19}}\ ev$

$W=0.1325\ eV$

Hence, The work done is $2.12\times10^{-20}\ J$ and 0.1325 eV.

Answer 2

Explanation:

Formula to calculate the electric field of the sheet is as follows.

          E = $\frac{\sigma}{2 \epsilon_{o}}$

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 = $\frac{\sigma}{2 \epsilon_{o}}$

                 = $\frac{4.60 \times 10^{-12}C/m^{2}}{2 \times 8.854 \times 10^{-12}C^{2}/N m^{2}}$

                 = 0.259 N/C

Now, force will be calculated as follows.

                 F = Eq

                    = $0.259 N/C \times 1.6 \times 10^{-19} C$

                    = $0.415 \times 10^{-19} N$

Now, work done will be as follows.

                    W = Fs

                        = $0.415 \times 10^{-19} N \times 0.51 m$

                        = $2.12 \times 10^{-20} J$

Thus, we can conclude that work done on the electron by the electric field of the sheet is $2.12 \times 10^{-20} J$.