.-45 points SerCP11 16.P006. My Notes Ask Your Teacher A point charge q = +37.0
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Question
.-45 points SerCP11 16.P006. My Notes Ask Your Teacher A point charge q = +37.0 pC moves from A to B separated by a distance d = 0.187 m in the presence of an external electric field E of magnitude 270 N/C directed toward the right as in the following figure. (a) Find the electric force exerted on the charge. magnitude direction (b) Find the work done by the electric force. (c) Find the change in the electric potential energy of the charge. (d) Find the potential difference between A and B. Need Help? ResdExplanation / Answer
a) The force (F) on a charge (q) in Electric Field (E) is q*E
F= 37*10-6 * 270 = 9.99*10-3 N (TOWARDS RIGHT)
b) Work is defined as F*d*cos(X) ; X is the angle between F and d
Here, X=0 , this implies, W=F*d=9.99*10-3 *0.187 = 1.8681*10-3 Joules
c) The electric field has done this work by consuming up some electric potential energy. Hence the electric potential energy now has reduced. Therefore the change in electric energy is negative (has got lowered as compared to before) and is equal to work done as no energy is lost. Hence, del.E= -1.8681*10-3 Joules
d) By looking at the direction of the electric field, we can say that A is at higher potential than B.Therefore VB - VA will be negative.
del.V = - (W/q) = - 50.4891 Volts
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