Answer
The electric potential at $P$ is $V\left( P \right) = 22743.1$ volts.
Work Step by Step
Let the circle be located at the first quadrant on the $xy$-plane. So, it can be parametrized by
${\bf{r}}\left( t \right) = \left( {4\cos t,4\sin t,0} \right)$, ${\ \ \ }$ for $0 \le t \le \pi /2$
Thus,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( { - 4\sin t,4\cos t,0} \right)\cdot\left( { - 4\sin t,4\cos t,0} \right)} $
$ = \sqrt {16{{\sin }^2}t + 16{{\cos }^2}t} = \sqrt {16} = 4$
$\delta \left( {{\bf{r}}\left( t \right)} \right) = 16\cos t\sin t$
The distance ${d_P}$ from $P = \left( {0,0,12} \right)$ to a point $\left( {x,y,0} \right)$ on the quarter circle is
${d_P} = \sqrt {{4^2} + {{12}^2}} = \sqrt {160} = 4\sqrt {10} $
Using Eq. (6), the electric potential at $P$ is given by
$V\left( P \right) = k\mathop \smallint \limits_C^{} \frac{{\delta \left( {x,y,z} \right){\rm{d}}s}}{{{d_P}}} = \frac{k}{{4\sqrt {10} }}\mathop \smallint \limits_0^{\pi /2} \delta \left( {x,y,z} \right){\rm{d}}s$
Using Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$,
where in this case $f\left( {x,y,z} \right) = \delta \left( {x,y,z} \right) = xy$, we obtain
$V\left( P \right) = \frac{k}{{4\sqrt {10} }}\mathop \smallint \limits_0^{\pi /2} \left( {16\cos t\sin t} \right)\cdot4{\rm{d}}t$
$V\left( P \right) = \frac{{8k}}{{\sqrt {10} }}\mathop \smallint \limits_0^{\pi /2} \sin 2t{\rm{d}}t = \frac{{4k}}{{\sqrt {10} }}\left( { - \cos 2t|_0^{\pi /2}} \right)$
$ = \frac{{4k}}{{\sqrt {10} }}\left( {1 + 1} \right) = \frac{{8k}}{{\sqrt {10} }}$
With the charge density in units of ${10^{ - 6}}$ and $k = 8.99 \times {10^9}$, we obtain
$V\left( P \right) = \frac{8}{{\sqrt {10} }} \times {10^{ - 6}} \times 8.99 \times {10^9} = 22743.1$
So, the electric potential at $P$ is $V\left( P \right) = 22743.1$ volts.
