Some equipotential surfaces are shown in figure(29.E3) What can you say about the magnitude and the direction of the electric field?

Explanation

We know, that the electric field is always perpendicular to equipotential surfac. So, making electric field lines perpendicular to the surface, we find that these lines are originating from the centre. So, the field is similar to that due to a point charge placed at the centre. So, comparing the given potentials with that due to point charge, we have,
`V=(KQ)/r" "implies" "KQ=V_(A)r_(A)=V_(B)r_(B)=V_(C)r_(C)=6 V-m`
Hence, electric field at distance r can be given by
`E=(KQ)/r^(2)=6/r^(2) V//m`
As the electric field lines are directed towards the decreasing potential. So, electric field is along radially outward direction.