A point charge q +45.0 HC moves from A to B separated by a distance d 0.196 m in

Question

A point charge q +45.0 HC moves from A to B separated by a distance d 0.196 m in the presence of an external electric field E of magnitude 260 N/C directed toward the right as in the following figure (a) Find the electric force exerted on the charge magnitude 1.17e-2 N direction toward the right (b) Find the work done by the electric force. 2.29e-3 (c) Find the change in the electric potential energy of the charge (d) Find the potential difference between A and B VB VA 50.96 The response you submitted has the wrong sign. V

Explanation / Answer

charge q=45 uC=45*10^-6 C

separation d=0.196 m

electric field E=260 N/C

a)

eletric force F=q*E

=45*10^-6*260

=0.0117 N

=1.17*10^-2 N

b)

workdone, W=F*d

=1.17*10^-2*0.196

=0.00229 J

=2.29*10^-3 J

c)

U=-W

=-2.29*10^-3 J

d)

potential difference VB-BA=U/q

=(-2.29*10^-3)/(45*10^-6)

=-50.88 V

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