# Relationship between charge and electric potential

"At any point in an electric field the electric potential is the amount of electric potential energy divided by the amount of charge at that point. It takes the charge quantity out of the equation and leaves us with an idea of how much potential energy specific areas of the electric field may provide."

My question is, why do we need to take the charge quantity out of the equation? It seems like dividing one by the other would give us some sort of rate (i.e. just like dividing distance by time gives us a rate of speed). Is that what we're trying to achieve by performing division here? Or is it closer to dividing in order to reduce a fraction, i.e. 5/15 is equal to 1/3?

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I think it may be because you need a test charge. The test charge's magnitude would influence the force experienced by the test charge, so you need to decouple the test charge from the equation. –  HL-SDK Feb 10 at 19:25
I think an analogy may be measuring gravitational force at a point by measuring the weight of an object and dividing that by its mass. 1 lb / 0.45kg ~~ 9.8m/s^2 –  HL-SDK Feb 10 at 19:27
This analogy was helpful, once I read that I was able to understand the answer below much more easily. Thanks for that. –  agentutah Feb 10 at 20:19
@HL-SDK that gives you acceleration due to gravity, or equivalently gravitational field strength. This is not analogous to electric potential, electrical potential energy, or electric charge. The electric analogy would be electric field strength, with units volts per meter. –  Phil Frost Feb 10 at 20:42

We want to find the potential at a point because if we have it we can compute the energy any particle of charge Q will have at that point by just multiplying Q by the potential V.

So the idea is to take a test charge with charge x Coulombs and place it at a point of unknown potential, we then measure the energy(E) exerted on that charge when placed at the point and compute the potential V as

$$V = \frac{E}{x}$$

This means that if I have any other charge (lets say Q), the energy exerted on it if it is placed at that point will be

$$E_q = QV = Q\frac{E}{x}$$

So we now know what the potential at that point is and we can now compute the energy a particle of any charge Q placed at that point will have without having to physically put that charge there.This is why the potential is the most important quantity to know, if you have it you can easily compute other quantities.

And to answer your other question, Yes, potential is a rate, it is joules per coulomb.

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At any point in an electric field the electric potential is the amount of electric potential energy divided by the amount of charge at that point.

Break it down, and consider the units for each quantity:

• the electric potential (volt, V)
• is the amount of electric potential energy (joule, J)
• divided by the amount of charge at that point (coulomb, C)

Mathematically:

$$\text{the electric potential} = \frac{\text{the amount of electric potential energy}}{\text{the amount of charge at that point}} \\$$

or:

$$\mathrm V = \frac{\mathrm J}{\mathrm C}$$

That is, the the definition of the volt.

It might make more sense to think of a rearrangement of this relationship:

At any point in an electric field, the amount of electric potential energy is the electric potential, multiplied by the amount of charge at that point.

$$\text{electric potential energy} = \text{electric potential} \cdot \text{charge at that point}$$

$$\mathrm J = \mathrm V \mathrm C$$

That is, voltage gives you an idea of how much potential energy you could have at some point, independent of how much charge you put there. Of course, if you are moving around a significant amount of charge, you are going to change the field, and thus the potential at each point.

Analogously, gravitational potential is measured in J/kg. Height times acceleration due to gravity yields gravitational potential, so if we assume a constant acceleration to gravity (such as approximately true for Earth's surface), then we can think about gravitational potential as height. Lifting a mass to some potential (height) takes some work, lifting twice that mass to the same potential takes twice as much work.

For another explanation, see this previous answer of mine.

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