# Voltage and Energy and lack of directionality

## Main Question or Discussion Point

To my understanding electric potential is the measure of energy per charge.
Now what I don't understand is that the voltage at the point P will be the same if there is a positive charge Q at A and a positive charge Q at B, or if there is a positive charge 2Q at A. I don't get how a test charge at point P should have the same amount of energy in both cases. Help.

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Hootenanny
Staff Emeritus
Gold Member
You are indeed correct that whether we have two equal charges, Q, at A and B; or a charge 2Q at A or B, the electric potential at P will be the same. It is important to emphasise that the electric potential is a measures of energy. Since the potential energy of a test charge only depends on the distance from a point charge, it isn't unreasonable that P should have the same potential energy in these two cases.

However, this does not mean that the force acting on a test charge at P would be the same. The force per unit charge at P is related to the gradient (or slope) of the potential at P, i.e. the force depends on how the voltage changes at P. So, although the value of the voltage might be the same at P, its slope may be different. Let me show you what I mean with a few plots.

This first image is when there are two equal charges, Q, located and A and B and shows the voltage (with zero at infinity).

The second is when there is a single charge, 2Q, located at A.

Notice that in both figures, you can choose any point along the line x=0, and the voltage will be equal. However, what is different is the gradient (or slope) of the voltage field. It is this that determines the force acting on a test charge. So whilst the force acting on a test charge is not the same in the two cases, the energy is.

Remember that the electric potential energy is the work done bringing a test charge from infinity, to a point in space. In this case, to a distance "r" from two charges of magnitude Q, or a distance "r" from a single charge of magnitude 2Q.

I hope that this helps.

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Well said.

Ok, that makes a lot of sense. Thank you