Answer
a) parallel
b) increases
c) 0.3mm
Work Step by Step
(a) It is given that $E=6500N/C$ and points in the negative x direction, thus the equipotential surfaces are perpendicular to the electric field; that is, the orientation of the equipotential surfaces is parallel to yz-plane.
(b) If we move in the positive x-direction -- that is, in the opposite direction to the electric field -- then the electric potential increases because the electric field points in the decreasing electric potential.
(c) The required distance can be determined as follows:
$\Delta S=\frac{\Delta V}{E}$
We plug in the known values to obtain:
$\Delta S=\frac{2V}{6500N/C}=3.077\times 10^{-4}m$
