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Could someone please describe the differences between the uses each of these potential equations:

Potential due to a point charge:

$V = \frac{k \cdot q}{r} - \frac{k \cdot q}{{r}_{\text{ref}}}$


Coulomb's potential (why are we missing the $r_{\text{ref}}$???):

$V = \frac{k \cdot q}{r}$


Electrostatic potential energy of a two-charge system:

$U = q_1 \cdot V = q_1 \cdot \frac{k \cdot q_2}{r} = \frac{k \cdot q_1 \cdot q_2}{r}$

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2 Answers

up vote 1 down vote accepted

The second and the third formulas are pretty much the same, the first formula differs from the second one just by a constant, so it provides the same electric field. The second formula uses a rather standard choice of the constant, such that the potential at the infinity is zero.

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I think your confusion may be addressed by an answer to a related question that I posted a while back. (And Lubos' answer too, probably) Basically, the reason is that potential is only defined relative to a reference point. So in a sense, the real, most general formula for potential is

$$V = -\frac{kq}{r} + \frac{kq}{r_\text{ref}}$$

using your notation.

However, it's a standard convention to choose $r_\text{ref} = \infty$ when it's possible to do so (which is when the charge distribution doesn't extend to infinity). This choice is convenient because it makes the second term equal to zero, and then you can write

$$V = -\frac{kq}{r}$$

It actually turns out not to matter what reference point you use, because potential enters into all physics formulas as a difference: you only ever use things like $V(r_2) - V(r_1)$, not $V(r_1)$ on its own. When you compute those differences, the term corresponding to the reference potential cancels out, even if it wasn't equal to zero. So basically, formula 2 is a convenient special case of formula 1.

The third formula in your question is something different. It's a formula for $U$, the potential energy, not the potential $V$. The difference is that $U$ depends on two charges, which are usually labeled the source charge and the test charge, whereas $V$ only depends on the source charge. $U$ is the energy of the combined system; $V$ is the energy per unit test charge. It's useful because you can calculate it based on only the source charge distribution, and then you can say something about the behavior of any test charge you put in, even without knowing in advance exactly how much the test charge is. For example, in a circuit with a battery, you know that the conducting electrons in the circuit will move from the negative terminal to the positive terminal, regardless of how many electrons there are (which you would need to know to calculate $U$).

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