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Edit View History Bookmarks People window Help Sun 615 PM Secure https session masteringphysics.com/myctitemview?assignmentProblemID 76420825&offset-next; Spring 2017 PHYS193 21197 L02 Item 5 tem 5 Part A An electron released from rest at a distanoe of How much work isdone on electron by the electric fold ofthe sheetasthe electron moves hom 0510 m from alarge insulating sheet of charge initalpositon to a point 700 102 mfrom the sheet? Bat has uniform surface charge density 30-10 12c/m Express your answer to three signifcant figures and include the appropriate units. O 63 W. Value Units Mr Answers give up Part B What is the speed of electron when it is 700 10 m from the sheet? Express your answer to three significant figures and include the appropriate units. Value Units 3 r 4 5 o 6 1 7 8 A 9

Explanation / Answer

Given

   electron released from rest

initially at a position x1 = 0.510 m from a uninformy charged sheet with surface charge density

   sigma = 3.00*10^-12 C/m2

and work done to bring the electron to a position x2 = 0.07 m from the sheet is = ?

we know that V = W/q

       W = V*q
and electric field of a sheet is E = sigma / 2 epsilon not

               E = 3.00*10^-12/(2*8.854*10^-12) N/C

               E = 0.16941495 N/C
Part A

and V = E*d

V1 = 0.16941495*0.510 V = 0.0864016245 v

and V2 = E*d1 = 0.16941495*0.07 V = 0.0118590465 V

   the work done W = (v2-v1)q = (0.0118590465 -0.0864016245)(1.6*10^-19) J = (-1.192681248)*10^-20 J

Part B

work done = change in kinetic energy

(-1.192681248)*10^-20 = 0.5(9.10*10^-31)(v)^2

v = 1.4733227*10^-25 m/s

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