Use the linear charge density $\displaystyle \lambda(x)=bx$ and length element $\displaystyle dx$, where each point is $\displaystyle \sqrt{x^2+a^2}$ from the point $\displaystyle (0,a)$. The potential is therefore

$\displaystyle V=\frac{1}{4\pi\epsilon_0}\int_{0}^{L}\frac{bx}{\sqrt{x^2+a^2}}dx$

$\displaystyle = \frac{b}{4\pi\epsilon_0}\left [ \sqrt{x^2+a^2}\right ]_{0}^{L}$

$\displaystyle =\frac{b}{4\pi\epsilon_0}\left (\sqrt{L^2+a^2}-a \right )$

Use the linear charge density $\displaystyle \frac{Q}{2\pi R}$ and length element $\displaystyle Rd\theta$. The distance from each point on the ring to the point on the axis is $\displaystyle \sqrt{R^2+a^2}$. Lastly, integrate over $\displaystyle \theta$ from $\displaystyle 0$ to $\displaystyle 2\pi$ to obtain

$\displaystyle V=\frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}\left (\frac{Q}{2\pi R} \right )\frac{1}{\sqrt{R^2+a^2}}Rd\theta$

$\displaystyle =\frac{Q}{4\pi\epsilon_0\sqrt{R^2+a^2}}$

Calculate the potential due to one side of the bar, and then multiply this by $\displaystyle 4$ to get the total potential from all four sides. Orient the bar along the x-axis such that its endpoints are at $\displaystyle \pm\frac{a}{2}$, and use the linear charge density $\displaystyle \frac{Q}{4a}$. The potential is therefore

$\displaystyle V=4\times\left (\frac{1}{4\pi\epsilon_{0}}\int_{-a/2}^{a/2} \left (\frac{Q}{4a} \right )\frac{1}{\sqrt{x^2+\frac{a^2}{4}}}dx\right )$

$\displaystyle =\frac{Q}{4\pi\epsilon_{0}a}\int_{-a/2}^{a/2} \frac{1}{\sqrt{x^2+\frac{a^2}{4}}}dx$

$\displaystyle =\frac{Q}{4\pi\epsilon_{0}a}\left[ \ln\left (\sqrt{x^2+\frac{a^2}{4}} +x\right )\right]_{-a/2}^{a/2}$

$\displaystyle =\frac{Q}{4\pi\epsilon_0a}\ln\left (\frac{\sqrt{2}+1}{\sqrt{2}-1} \right )$

$\displaystyle =\frac{Q}{2\pi\epsilon_0a}\ln(1+\sqrt{2})$

According to the shell theorem, the total electric field at the center point of a charged spherical shell is always zero. At this point, any electric field lines will result in symmetry, canceling each other out and creating a net field of zero at that point.

The original charge Q is negative, as indicated by the direction of the electric field lines in the diagram. A negative point charge of any value placed at point A will cause both charges to feel a mutually repelling force on each other. Therefore, the second charge will be repelled by the original negative charge with a force pointing radially along a line connecting the center of Q and the point A, resulting in free movement toward C. Additionally, point B is on a line of equipotential to point A; negative point charges will move from regions of lower electric potential to higher electric potential (against the direction of the electric field lines), so point B is not a viable answer.