A uniform electric field of magnitude 423 N/C pointing in the positive x-directi

Question

A uniform electric field of magnitude 423 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.10 cm.
(a) What is the work done by the field on the electron? J
(b) What is the change in potential energy associated with the electron? J (c) What is the velocity of the electron? magnitude m/s
direction A uniform electric field of magnitude 423 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.10 cm.
(a) What is the work done by the field on the electron? J
(b) What is the change in potential energy associated with the electron? J (c) What is the velocity of the electron? magnitude m/s
direction A uniform electric field of magnitude 423 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.10 cm.
(a) What is the work done by the field on the electron? J
(b) What is the change in potential energy associated with the electron? J (c) What is the velocity of the electron? magnitude m/s
direction

Explanation / Answer

Given:-

E=423 N/C pointing in the positive x-direction

d=3.10*10-2 m

(a) work done by the electron, W= F*d= qE*d= 1.6*10-19*423*3.10*10-2 = 2.098*103*10-21= 2.098*10 -18 J

(b)

the potential energy of a charged particle in a uniform electric field is:

U = qEd where

Hence change in potential energy=work done=2.098*10 -18 J

[c] Applying conservation of energy

Ke of electron= Pe

1/2mv2= 2.098*10 -18

1/2*9.1*10-31 v2=2.098*10 -18

v2=2.098*10 -18 / 4.55*10-31

v= 2.15*106 m/s,pointing in the positive x-direction

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