(a) A proton is released from rest at x = -2.00 cm in aconstant electric field w

Question

(a) A proton is released from rest at x = -2.00 cm in aconstant electric field with magnitude 1.50 x103 N/C pointing in the positive x-direction.Assuming an initial speed of zero, find the speed of a proton atx = 0.0500 m with a potential energy of -1.65 x 10-17 J.

(b) An electron is now fired in the same directionfrom the same position. Find the initial speed of the electron (atx = -2.00) given that its speed has fallen by half when itreaches x = 0.120 m, a change in potential energy of3.33 x 10-17J.

C) The electron in part (b) travels fromx = 0.120 m to x = -0.180 m within the constantelectric field. If there's a change in electric potential energy of-8.90 x 10-17J as it goes from x = 0.120 m to x = -0.180 m,find the electron's speed at x = -0.180 m.

Explanation / Answer

C) The electron in part (b) travels fromx = 0.120 m to x = -0.180 m within the constantelectric field. If there's a change in electric potential energy of-8.90 x 10-17J as it goes from x = 0.120 m to x = -0.180 m,find the electron's speed at x = -0.180 m. change in potential energy =kinetic energy=1/2mv*v 8.9*10-17=1/2*9.1*10-31*v*v v=2*8.9*10-17/9.1*10-31=1.3985*107m/sec

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