# Physics with Calculus/Electromagnetism/Voltage

To develop the electric potential (or voltage), begin with a single point charge, Q, at the origin, of charge q. The electric field is

$E={\frac {kQ}{r^{2}}}{\hat {r}}$ .

The curl of this is easily shown to be zero, and by a well-known theorem of vector calculus, if the curl of a function vanishes over a region of space, then that function may be written as the gradient of a scaler. Hence, there exists a scaler field, $\phi$ , such that

$E(\mathbf {r} )=-\nabla \phi (\mathbf {r} )$ .

We next find the work done by the field as it moves a test particle, q, from point $\mathbf {a}$ to $\mathbf {b}$ . y the force field, $\mathbf {F} =q\mathbf {E}$ . If you do not know vector calculus, you can skip this paragraph and only remember the result, but I suggest you learn vector calculus (divergence, gradient, and curl) because all of electromagnetism is written with vector calculus notations. The curl of E is zero (you may do this out). This means that E can be written as the gradient of some scalar function, let us call it $\phi$ . By the fundamental theorem of calculus (for gradients),

$\int q\nabla \phi (x,y,z)\cdot d{\mathbf {r} }=q\phi +C$ ,

where C is an arbitrary constant. This states that the work done by the electric force is simply the change in $\phi$ times the charge. This should sound familiar -- $q\phi$ is the potential energy! We specify $\phi$ at one point, and it determines $\phi$ uniquely, which is exactly the same as selecting a zero for gravitational potential energy. Conventionally, we say $\phi (\inf )=0$ , but this is not always true if the charge extends to infinity. However, this only happens in bogus textbook problems, so in real life, it is always safe to say $\phi$ drops off to zero at infinity. We are allowed to select a reference point, because it is only the difference in potential energy we care about; there is no physical way to determine C, so we don't care what it is.

To generalize to anything more than a point charge is easy. It is just the superposition of many point charges, and our result holds true for any distribution.

Now, it is conventional to say that $\nabla \phi$ is negative E, because it means things travel to lower potentials, and it cancels out a minus sign somewhere else. Just watch out for this, and keep track of your minus signs.

To apply this to Coulomb's law, we just take the integral of the electric field and get the potential. We stick on a minus sign for the convetion and we have,

$\phi _{pointsource}={\frac {kq}{r}}$ .

How simple.

There are two main advantages to using potential. The first is that it will come in handy later when we want to solve Maxwell's equations, but the second reason is that it is a scalar, not a vector. It is much easier to only remember one component instead of three. You find the potential of a charge distribution, then take finding the E field is as easy as taking a derivative.

Potential is an interesting thing, because it is a scalar function, yet it uniquely represents a vector function, E, which has three times as many numbers! Well, it just so happens that the three components of an electric field are not independent, because they are radially symmetric around a point charge, so the curl is zero. This constraint is just enough to tell us that we only need one number to represent all three components of E.

I would like to point out that the potential only can be used (without modification) when the charges are not moving. If they move, then if a particle goes around in a circle, it may end up with more energy than it had before!