Q. Figure  shows some equipotential surfaces. What can you say about the magnitude and direction of the electric field.


angle b/w electric field and dx is 90o+30o
change in first and second potential surface=dv=10V
so,E.dx= -dV
Edxcos(90o+30o)= -dv
E(10*10-2)cos 120o= -10
the electric field is making an angle of 120o with the x-axis

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