How to Calculate Electric Potential at Point P in a Triangle with Charges?

AI Thread Summary
To calculate the electric potential at point P in a triangle with charges, the formula V(r) = kq/r is used, where k is the Coulomb's constant, q is the charge, and r is the distance from the charge to point P. The sign of each charge must be considered when summing the voltages, as electric potential is a scalar quantity. For example, with charges +3q and -3q positioned horizontally relative to point P, their contributions can cancel each other out, resulting in zero potential at that point. However, the presence of a test charge would indicate a non-zero electric potential energy due to the forces acting on it. Ultimately, the distances from each charge to point P are crucial in determining the total electric potential.
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Homework Statement



http://s62.photobucket.com/albums/h116/pepsi_in_a_can/?action=view&current=diagram.jpg

Each side of the triangle has a given length "d." Calculate the electric potential at point P.



Homework Equations



V(\vec{r}) = \frac{kq}{r}

I believe this is for the elec. potential at some distance r for a point charge q, but I'm not completely sure.

The Attempt at a Solution



My main problem is whether or not to take the sign of each charge into account when I'm summing the voltage. For instance, when considering the charges horizontal to the point P, +3q and - 3q, if I take the signs into account, I end up with zero voltage. I've read that the idea of superposition applies, but the voltage is a scalar, and needs to be handled differently.

A push in the right direction would be much appreciated.
 
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Calculate the electric potential due to each charge taking into account its sign.
and add them.
when considering the charges horizontal to the point P, +3q and - 3q, if I take the signs into account, I end up with zero voltage It is right.
 
Ok, let's just say that I had the +3q and -3q charges, and the point P midway between them. If I say that the electric potential at P is +3e + (-3q) = 0, then I have 0 potential at that point. This seems strange to me. If I placed a positive test charge at point P, it would move away from +3q and move toward -3q, which infers that it has non-zero potential enery, and thus non-zero electric potential.
 
You're correct in being suspicious of that hotcommodity

Remember one way to think of electric field is force per unit charge

So if you put a unit charge there and there's a force(like you said), there's clearly electric potential there
 
So how do I add the voltages? I assume that charge plays role in determining whether there's positive or negative voltage at anyone point, but I'm not sure how to apply that concept. Additionally I'm wondering if any angles need to be taken into account when adding the voltages.
 
For the question you asked, with just the two charges in a line, then no, no need to worry about angles

Remember what r is, it's the distance between the source(the charge)and the point(in this case point P), so assuming everything is lined up on the x axis

Let's say the -3q is on the left at x=-3 and the positive 3q is at x=3 and point P is at the origin

You'd have the electric potential at P because of the left charge be kq/r, q=-3q and r = -3...

See where this is going?
 
So given those values of r, I'd have an electric potential of 2q at P?
 
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