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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?

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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.

This assumption is true for system of two charges of which one of them is +ve test charge but if there are more charges in the system you can still apply that logic but taking two charges at a time but not for system of charges altogether and for that you need to know about superposition theorem.

In Superposition theorem(in this case) ,you can remove all the charges from system of charges except test and one charge(Q1) from the system and calculate potential due to that one remaining charge (and you can apply attraction-repulsion logic here because there is only two charges .

And repeat above process again but with test charge and other charge (Q2) of the system and in this manner calculate potential due to all the charges and then add them ,the net potential will the potential due to system of charges

Thus I would assume using a signed distance (1 and −1 in this case) to be the correct approach

This is wrong ! Distance (r) is always taken as positive but sign of charges (Q1,Q2 etc ) is important .

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

Why you expecting that potential should cancel each other although I explained above why this is not the case!

previous logic of the sign of the potential signaling repulsion or attraction would not work anymore.

As I explained earlier it still works if you use Superposition theorem but not for system of charges all taken together

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  • \$\begingroup\$ Correct me if I'm wrong but isn't my calculation (the summation) that I included in my question exactly applying the superposition theorem that you mention? Why am I assuming that the charges should cancel: Let's say Q is positive, thus it will repel the test-charge of 1C. One point-charge Q1 lies at the left of the origin, it will push a test-charge on the origin to the right. The other point-charge Q2 lies at the right of the origin, it will push a test-charge on the origin to the left. This is why I assume they cancel each other out. \$\endgroup\$ – LeonTheProfessional Jan 6 at 10:51
  • \$\begingroup\$ @LeonTheProfessional Yes! you applied superposition but as you later used wrong sign of r (distance ) so it was necessary to tell you how to use correctly!I think you mixing net force acting on charge is zero with potential being zero , net force is zero at origin because net electric field is zero not because net potential is zero . \$\endgroup\$ – user215805 Jan 6 at 11:07
  • \$\begingroup\$ Yes, very likely I misunderstand the concept of potential, but: Isn't the potential saying that a charge placed at a point of certain potential has a certain ability to perform work? So how is a charge of 1C placed at the origin able to perform work in the model outlined above? \$\endgroup\$ – LeonTheProfessional Jan 6 at 11:54
  • \$\begingroup\$ @LeonTheProfessional , yeah it seems like if we displace test charge slightly from origin and towards x-axis and due to more repulsion it come back to its original position with some velocity or in other words it performs SHM along X-axis ,so it seems like it should be stable equilibrium point(Not really ) ?but what about other directions let's say we displace slightly along at some angle from X-axis ,now what will happen ? This time charge leave its position and moves such that it minimize potential energy of the system which is zero (in this case) at infinity ! \$\endgroup\$ – user215805 Jan 6 at 13:15
  • \$\begingroup\$ And hence we can say field can do +ve work on it \$\endgroup\$ – user215805 Jan 6 at 13:15

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