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
E=200V/m
the electric field is making an angle of 120o with the x-axis
change in first and second potential surface=dv=10V
so,E.dx= -dV
Edxcos(90o+30o)= -dv
E(10*10-2)cos 120o= -10
E=200V/m
the electric field is making an angle of 120o with the x-axis