\Part{b}
The geometry is flipped for $B$, so we have
\begin{equation}
- V_A = k\frac{Q}{d\sqrt{2}} + k\frac{2Q}{d}
+ V_B = k\frac{Q}{d\sqrt{2}} + k\frac{2Q}{d}
= k\frac{Q}{d}\p({2+\frac{1}{\sqrt{2}}})
- = \ans{6.08\E{-15}\U{V}} = \ans{6.08\U{kV}} \;,
+ = \ans{6.08\E{3}\U{V}} = \ans{6.08\U{kV}} \;,
\end{equation}
\Part{c}