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?

Explanation:

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

          E =

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 =

                 =

                 = 0.259 N/C

Now, force will be calculated as follows.

                 F = Eq

                    =

                    =

Now, work done will be as follows.

                    W = Fs

                        =

                        =

Thus, we can conclude that work done on the electron by the electric field of the sheet is .

The work done is and 0.1325 eV.

Explanation:

Given that,

Distance = 0.570 m

Surface charge density

Using formula of electric field

Using formula of electrostatic force ,

(A). We need to calculate the work done

Using formula of work done

Put the value into the formula

Hence, The work done is and 0.1325 eV.

a. are