The distance between the points A and B on two equipotential lines with V1=5.6 V and V2=1.7 V is 3.5 cm. What is the average electric field at the midpoint C expressed to one decimal place?
Homework Answers
as electric field =-change in potential/distance
potential at C=(V1+V2)/2=(5.6+1.7)/2=3.65 volts
distance between A and C=3.5 cm/2=1.75 cm=0.0175 m
so electric field =-(3.65-5.6)/0.0175 V/m=111.4285 V/m
in one place after decimal, electric field=111.4 V/m
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