A point charge q = +41.0 µC moves from A to B separated by a distance d = 0.174
ID: 2048128 • Letter: A
Question
A point charge q = +41.0 µC moves from A to B separated by a distance d = 0.174 m in the presence of an external electric field of magnitude 291 N/C directed toward the right as in the following figure.
(a) Find the electric force exerted on the charge.
magnitude_____ N
direction_______---Select--- toward the right toward the left The magnitude is zero.
(b) Find the work done by the electric force.
______ J
(c) Find the change in the electric potential energy of the charge.
_________ J
(d) Find the potential difference between A and B.
VB - VA = _______ V
Explanation / Answer
(a)
F=qE
=41.0 µC*291 N/C=0.011931 N=11.931x10-3 N
AS,Electric field is directed towards right and the charge is positive ,So,it will feel force in the direction of Electric field.
Towards Right is the direction.
(b)
W=F.D=11.931x10-3 N*0.174 m =2.076x10-3 Joule
(c)
change in the electric potential energy of the charge=Work done =2.076x10-3 joule
(d)
VB - VA =E.d=291 N/C*0.174 m= 50.634 volt
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