given
charge q = 42*10^-9 C
distance d = 0.165 m
electric field E = 300 N/C
a) from the relation
electric force F = qE
F = 42*10^-9*300 = 1.26*10^-5 N
direction: in the direction of electric field that is towards right
b) work w = F*d = 1.26*10^-5*0.165 = 2.079*10^-6 J
c) from the relation work = change in electric potential energy = 2.079*10^-6 J
d) potential difference = work/charge = 2.079*10^-6/42*10^-9 = 49.5 V
-19 points SERCP11 16.1.OP.006.0/5 Submissions Used My Notes Ask Your Teacher 42.0 nC, that moves a...
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