Two point charges \(-q\) and \(+q\) are located at points \((0,0,-a)\) and $(0,0, a)\( respectively. The potential at a point \)(0,0, z)\( where \)z>a\( is \)\ldots \ldots$ (A) $\left[(2 \mathrm{q} a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}+a^{2}\right)\right\\}\right]$ (B) \(\left[\mathrm{q} /\left(4 \pi \epsilon_{0} \mathrm{a}\right)\right]\) (C) \(\left[\right.\) (qa) \(\left./\left(4 \pi \in_{0} z^{2}\right)\right]\) (D) $\left[(2 q a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}-a^{2}\right)\right\\}\right]$

Determine the distance between the point \((0,0,z)\) and each of the point charges \(-q\) and \(q\). These distances will be needed to calculate the electric potentials due to each point charge. Distance to \(-q\): \(r_{1}=|z-(-a)|=z+a\) Distance to \(+q\): \(r_{2}=|z-a|\)

Now, we will calculate the electric potential at point \((0,0,z)\) due to each point charge using the formula: \[V = \frac{Q}{4\pi\epsilon_{0}r}\] Electric potential due to \(-q\): \[V_{1} = \frac{-q}{4\pi\epsilon_{0}(z+a)}\] Electric potential due to \(q\): \[V_{2} = \frac{q}{4\pi\epsilon_{0}(z-a)}\]

Now, we will calculate the total electric potential at point \((0,0,z)\) by adding the electric potentials due to each point charge: \[V_{total} = V_{1} + V_{2} = \frac{-q}{4\pi\epsilon_{0}(z+a)} + \frac{q}{4\pi\epsilon_{0}(z-a)}\] By taking the common factor, we get: \[V_{total} = \frac{q(-1+1)}{4\pi\epsilon_{0}(z^{2}-a^{2})} = \frac{0}{4\pi\epsilon_{0}(z^{2}-a^{2})}\] Now, comparing this result to the given options, we find that none of the options match our result. Therefore, there is most likely an error in the given options, and the correct answer should be: \[V_{total} = 0\] This also makes physical sense, as the two charges are equal in magnitude but opposite in sign, so their potentials can cancel each other out at some points in space.