Answer
(a) The potential increase and it doesn't depend on the reference point.
(b) The distance between two surfaces is 0.29 cm
Work Step by Step
(a) The potential increase as the surface charges are negative where the motion is in the opposite direction of the electric field and it doesn't depend on the reference point.
(b) The electric field is related to the charge density by
\begin{gather}
E=\frac{\sigma}{ 2\epsilon_{o}}\\
E=\frac{-6 \times 10^{-9}\mathrm{~C/m^2}}{2 \times 8.85 \times 10^{-12}\,\text{F/m}}\\
E =-3.38 \times 10^{2} \mathrm{V} / \mathrm{m}
\end{gather}
The potential difference is related to the electric field and the distance between two surfaces $d$
\begin{equation}
V=E d\\
d=\frac{V}{|E|} \\
d=\frac{1 \mathrm{V}}{|-3.38 \times 10^{2} \mathrm{V} / \mathrm{m}|} \\
d =0.0029 \mathrm{m} \\
\boxed{d =0.29 \mathrm{cm}}
\end{equation}
