I am reading about the electric potential:
from wikipedia:
The electric potential (...) is the amount of work energy needed to move a unit of electric charge (a Coulomb) from a reference point to the specific point in an electric field ...
Consider then an electric potential due to a point-charge \$Q\$:
I would assume that the potential should be positive whenever our (unit)test-charge and the point-charge repel each other, and negative when they attract each other. Since our test-charge is fixed to be \$1C\$, we only need to consider the charge \$Q\$, and our electric potential will be proportional to this charge.
Let's say we consider the potential \$ \varphi_0 \$ at the origin. Whether \$Q\$ lies at \$\left(1,0,0 \right)\$, or at \$\left(-1,0,0\right)\$ shouldn't matter since we only consider attraction or repulsion. Also I see many formulas just considering \$r\$ which to me doesn't look like there is a sign involved.
Now here is where my problem starts: Consider now two point charges \$Q_1\$ at \$\left(-1,0,0 \right)\$ and \$Q_2\$ at \$\left(1,0,0 \right)\$. Let \$Q_1 = Q_2\$
Wikipedia also explains how to calculate the potential due to multiple point-charges, namely by just adding up the individual potentials due to one point charge.
Now if we consider again the potential \$\varphi_0\$ at the origin, we would have: $$ \varphi_0 = \frac{1}{4\pi\epsilon_0} \left(\frac{Q}{1} + \frac{Q}{1} \right) = \frac{2Q}{4\pi\epsilon_0} $$ Thus we have double the potential.
Now this is not what I expect. Since the point-charges lie on opposite sides of the origin, I would very much expect them to cancel each other out when calculating the potential at the origin. Thus I would assume using a signed distance (\$1\$ and \$-1\$ in this case) to be the correct approach, but then the previous logic of the sign of the potential signaling repulsion or attraction would not work anymore.
Where has my reasoning gone astray? Where am I making wrong assumptions? Where am I misreading/misunderstanding the wikipedia-explanations?