# Electric Potential between two points

## Homework Statement

Two equal point charges Q = 5.18 C are separated by a distance d = 2.00 m. (See figure.)

[URL]http://homework.phyast.pitt.edu/res/msu/nagytibo/Electromagnetism/Electrostatics/graphics/010a.gif[/URL]

Point A is halfway between the charges and point B is located 1.00 m to the right of the second charge. What is the electric potential at point A?

V = G*q / r

## The Attempt at a Solution

I thought the potential would be 0, since it's half-way between the two points, meaning that the Electric Field would be 0 there, but that wasn't the correct answer.

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Seems like you're getting the electric potential and the electric field confused.
If you had a single point charge what would the electric potential be at that point? Now if you added more charge would you expect the electric potential to increase or decrease?

If the charge was on the opposite side of the point, the electric field lines would be pointing in the opposite direction, so wouldn't the opposite E-field lines become negative for potential and cancel out the other charge?

kreil
Gold Member
Since the charges are the same sign the fields will add (if one was a negatively charged particle the fields would cancel)

So I just add the two electric potentials together instead of subtracting them?

Could you explain why they add together instead of canceling? If I released the particle at the center, there wouldn't be any net force exerted on it, and it would remain stationary.

I don't know of a great way to visualize it but think about the electric potential decreasing as it moves away from the point charge so you have only so much left as you move further away.

Each point is just another electric potential that is vectorially outward (assuming a positive charge). So if you added another set of electric potentials that overlap the first then via superposition the total electric potential would be this newly defined gradient.

I don't know if I was clear but I hope that helps.

Each point is just another electric potential that is vectorially outward (assuming a positive charge). So if you added another set of electric potentials that overlap the first then via superposition the total electric potential would be this newly defined gradient.
Electric potential is not a vector quantity.

and this the solution of your question too 123yt.

the potential due to 2 points is added algebraically as its a scalar