Answer
(a) Point a is at a higher electric potential than point b.
(b) We can rank the magnitudes of the potential differences from largest to smallest:
$\Delta V_{ef} \gt \Delta V_{cd} \gt \Delta V_{ab}$
(c) Surface 1 is an equipotential surface.
Surface 2 is not an equipotential surface.
Work Step by Step
(a) The electric field points from positions of higher potential to positions of lower potential. Therefore, point a is at a higher electric potential than point b.
(b) We can write an expression for the magnitude of the potential difference:
$\Delta V = \Delta x ~E$
Note that $\Delta V$ increases as $\Delta x$ increases.
We can rank the magnitudes of the potential differences from largest to smallest:
$\Delta V_{ef} \gt \Delta V_{cd} \gt \Delta V_{ab}$
(c) Surface 1 is perpendicular to the electric field. Therefore, surface 1 is an equipotential surface.
Surface 2 is not perpendicular to the electric field. Therefore, surface 2 is not an equipotential surface.
