Editing Classical electromagnetism (section)

==Electric field==
{{Main|Electric field}}

The electric field '''E''' is defined such that, on a stationary charge:

:<math>
\mathbf{F} = q_0 \mathbf{E}
</math>

where ''q''<sub>0</sub> is what is known as a test charge and {{math|'''F'''}} is the [[Electrostatic force|force]] on that charge. The size of the charge does not really matter, as long as it is small enough not to influence the electric field by its mere presence. What is plain from this definition, though, is that the unit of {{math|'''E'''}} is N/C ([[newton (unit)|newtons]] per [[coulomb]]). This unit is equal to V/m ([[volt]]s per meter); see below.

In electrostatics, where charges are not moving, around a distribution of point charges, the forces determined from [[Coulomb's law]] may be summed. The result after dividing by ''q''<sub>0</sub> is:

:<math>\mathbf{E(r)} = \frac{1}{4 \pi \varepsilon_0 } \sum_{i=1}^{n} \frac{q_i \left( \mathbf{r} - \mathbf{r}_i \right)} {\left| \mathbf{r} - \mathbf{r}_i \right|^3}</math>

where ''n'' is the number of charges, ''q<sub>i</sub>'' is the amount of charge associated with the ''i''th charge, '''r'''<sub>''i''</sub> is the position of the ''i''th charge, '''r''' is the position where the electric field is being determined, and ''ε''<sub>0</sub> is the [[electric constant]].

If the field is instead produced by a continuous distribution of charge, the summation becomes an integral:

:<math>\mathbf{E(r)} = \frac{1}{ 4 \pi \varepsilon_0 } \int \frac{\rho(\mathbf{r'}) \left( \mathbf{r} - \mathbf{r'} \right)} {\left| \mathbf{r} - \mathbf{r'} \right|^3} \mathrm{d^3}\mathbf{r'}</math>

where <math>\rho(\mathbf{r'})</math> is the [[charge density]] and <math>\mathbf{r}-\mathbf{r'}</math> is the vector that points from the volume element <math>\mathrm{d^3}\mathbf{r'}</math> to the point in space where '''E''' is being determined.

Both of the above equations are cumbersome, especially if one wants to determine '''E''' as a function of position. A scalar function called the [[electric potential]] can help. Electric potential, also called voltage (the units for which are the volt), is defined by the [[line integral]]

:<math>
\varphi \mathbf{(r)} = - \int_C \mathbf{E} \cdot \mathrm{d}\mathbf{l}
</math>

where <math>\varphi(\textbf{r})</math> is the electric potential, and ''C'' is the path over which the integral is being taken.

Unfortunately, this definition has a caveat. From [[Maxwell's equations]], it is clear that {{nowrap|∇ × '''E'''}} is not always zero, and hence the scalar potential alone is insufficient to define the electric field exactly. As a result, one must add a correction factor, which is generally done by subtracting the time derivative of the '''A''' vector potential described below. Whenever the charges are quasistatic, however, this condition will be essentially met.

From the definition of charge, one can easily show that the electric potential of a point charge as a function of position is:

:<math>
\varphi \mathbf{(r)} = \frac{1}{4 \pi \varepsilon_0 }
\sum_{i=1}^{n} \frac{q_i} {\left| \mathbf{r} - \mathbf{r}_i \right|}
</math>

where ''q'' is the point charge's charge, '''r''' is the position at which the potential is being determined, and '''r'''<sub>''i''</sub> is the position of each point charge. The potential for a continuous distribution of charge is:

:<math>
\varphi \mathbf{(r)} = \frac{1}{4 \pi \varepsilon_0}
\int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}\, \mathrm{d^3}\mathbf{r'}
</math>

where <math>\rho(\mathbf{r'})</math> is the charge density, and <math>\mathbf{r}-\mathbf{r'}</math> is the distance from the volume element <math>\mathrm{d^3}\mathbf{r'}</math> to point in space where ''φ'' is being determined.

The scalar ''φ'' will add to other potentials as a scalar. This makes it relatively easy to break complex problems down into simple parts and add their potentials. Taking the definition of ''φ'' backwards, we see that the electric field is just the negative gradient (the [[del]] operator) of the potential. Or:

:<math> \mathbf{E(r)} = -\nabla \varphi \mathbf{(r)} .</math>

From this formula it is clear that '''E''' can be expressed in V/m (volts per meter).