Answer
a). (1) higher
b). $-1.23\times10^{-5}V$
Work Step by Step
a). For a positive charge, the electric field points in radially outward direction.
The electric potential decreases in the direction of electric field. So a point far away from the positive charge will have lower electric potential than the point near the point charge.
So, (1) higher is the answer.
b). $q=5.5\times10^{-6}C$,
$r_{1}=20\times10^{-2}m$,
$r_{2}=40\times10^{-2}m$
Thus, $V=k\frac{q}{r}$
So, $\Delta V= kq(\frac{1}{r_{2}}-\frac{1}{r_{1}})=-1.23\times10^{-5}V$
