Question

(HW03-22.23) An electron is released from rest at a distance of 0.600 m from a large insulating sheet of charge that has uniform surface charge density 4.30×10^?12C/m2 .

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 3.00×10^?2 m from the sheet?

Express your answer to three significant figures and include the appropriate units.

B)

What is the speed of the electron when it is 3.00×10^?2 m from the sheet?

Express your answer to three significant figures and include the appropriate units.

Answer 1

(A) At first, find the field from the sheet of charge.

This is E = ? / (2?0) = (4.3 x 10^-12) / (2 x 8.854 x 10^-12) = 0.243 N/C

Now, find the force on the electron using F = qE = 1.6 x 10^-19 x 0.243 = 0.389 x 10^-19 N

And at last, find the work done on the electron using W = Fd where d = 0.600 - (3.00×10^?2) = 0.57 m

So, W = 0.389 x 10^-19 x 0.57 = 2.22 x 10^-20 J

(B) The work done will equal to the kinetic energy gained.

Therefore, we have -  
½mv² = 2.22 x 10^-20

=> 0.5 x 9.1 x 10^-31 x v^2 = 2.22 x 10^-20

=> v^2 = 0.4879 x 10^11 = 4.879 x 10^12

=> v = 2.21 x 10^6 m/s.