Energy and the Scalar Electric Potential

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Suppose we have region filled with a static electric field and we introduce a "test" charge $δ Q$ to the region. Somehow, we are capable of moving this test charge around. Since the charge feels a force due to the field, work is done as we move it around. The work done is simply

$W=-\int_C \textbf{F}\cdot\, d\textbf{l},$

Illustration of the path taken by the test charge.

where $\textbf{F}$ is the force experienced by the charge and $d\textbf{l}$ is the differential element of length along the contour C. The negative sign is chosen so that when the charge does work against the field, the work is positive and vice versa.

We know that the force on the charge is given by

$\textbf{F} = \delta Q \textbf{E},$

so we can write the work done (corresponding to the energy change of the field/test-charge system) as

$W = -\delta Q \int_C \textbf{E}\cdot\, d\textbf{l}.$

The work done by a charge moved about in a static electric field does not depend on the path taken by the charge, but only on the endpoints. Hence, the work done can be written as

$W = \delta Q (V_2 - V_1)\,$ .

The quantity between the parantheses is known as the potential difference between points 1 and 2. This potential has units of volts and is defined as

$V_2 - V_1 = -\int_C \textbf{E}\cdot \, d\textbf{l}.$

If the path C is a closed loop, the potential difference (and hence to total work done by the charge) will vanish. This is typical of static electric fields, which are known as conservative fields. (Static gravity is another conservative field. Lifting a weight and putting it back down in the same spot incurs no net energy expenditure, in an "ideal" world.)

Simply put, this means that any charged body moved through a potential difference causes a change in system energy. Moving from a lower to a higher potential means that the charge works against the field, i.e. we push the charge. Moving from a higher to a lower potential, the field does work on the charge, i.e. the charge pushes us.

1 Recovering electric field from a known potential
2 Relationship to Maxwell's Equations
3 Potentials derived from point charges
- 3.1 Single point charge
- 3.2 Dipole
4 Distributed charges
- 4.1 Example

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Recovering electric field from a known potential

In a static electric field, the force on the test charge can also be written in terms of the change in system energy. If we assume the starting potential V1 to be a constant and drop the subscript 2, the force is

$\textbf{F} = -\delta Q\, \nabla V.$

Hence, for the static field, we can write

$\textbf{E} = -\nabla V.$

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Relationship to Maxwell's Equations

The preceding discussion of scalar potential is valid only if

$\nabla\times\textbf{E} = 0$ .

That is to say, when there is no time variation, or $\partial\textbf{B}/\partial t = 0$ . If we move the test charge through any closed path $\partial S$ , the path defines a surface $S\,$ . Hence, it can be shown that

$\int_S \nabla\times\textbf{E}\cdot\hat{\textbf{n}}\, dS = \oint_{\partial S}\textbf{E}\cdot\, d\textbf{l} = 0$ ,

if we assume the convention that the enclosing boundary is oriented anticlockwise and the surface unit vector follows the "right-hand-rule" (the fingers of your right hand follow the anticlockwise sense of the enclosing curve and your thumb will point in the direction of $\hat{\textbf{n}}\,$ .

In the case where the curl of the electric field vanishes, it is possible to define all the behaviour of the electric field using the scalar electric potential V. This is a powerful mathematical trick used to simplify electric field problems at DC (or where the frequency is low, i.e. 50-60Hz). This is because any field defined by a gradient of a scalar potential will always have a zero curl, viz.

$\nabla\times\textbf{E} = \nabla\times(-\nabla\, V) = 0$ .

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Potentials derived from point charges

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Single point charge

For the single point charge $Q$ in empty space, we can make the assumption that the potential infinitely far from the charge is zero. In this case, the potential is

$V(\textbf{r}) = \frac{Q}{4\pi \epsilon_0 \vert r\vert}$

where $\textbf{r}$ is a vector pointing from the charge to the observation point. By taking the gradient of this expression, the electric field of the single point charge easily found.

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Dipole

Let us revisit the interesting case of the dipole. Remember that the dipole consists of two point charges of equal magnitude but opposite sign spaced by a distance of $2 b$ . Since the electric potential is a linear function of charge, the principle of superposition applies and we can write the dipole potential as the sum of the potentials of the two point charges:

$V = \frac{Q}{4\pi \epsilon_0}\left (\frac{1}{r_1} - \frac{1}{r_2} \right ).$

Geometry of the electric dipole

Since the position vector stretches from the midpoint of the dipole to the observation point, it is straightforward to use the Law of Cosines to put $r_1\,$ and $r_2\,$ in terms of the position vector coordinates $r\,$ and $\theta\,$ :

$r_1 = \sqrt{r^2\left (1+\frac{b^2}{r^2}-2\frac{b}{r}\cos\theta\right )}$

$r_2 = \sqrt{r^2\left (1+\frac{b^2}{r^2}+2\frac{b}{r}\cos\theta\right )}$

If the observation point distance is much larger than the dipole separation (i.e. $r > > b$ ), then the two distances $r_1\,$ and $r_2\,$ can be approximated as

$r_1 \approx r-b\cos\theta,$

$r_2 \approx r+b\cos\theta.$

Inserting these into the initial expression for dipole potential and rearranging the fractions a bit yields

$V = \frac{Qb}{2\pi\epsilon_0 r^2}\cos\theta$

for the dipole potential. Note that this is an approximation and is valid only if $r\,$ is generally greater than ten times the dipole spacing $b\,$ . Also, the product $2Qb\,$ is called the dipole moment. The dipole moment is a vector. In our example, there only the z component because the charges are lined up along the z-axis. For a dipole moment that points in an arbitrary direction and sits at an arbitrary position, the potential can be written in terms of the vector dipole moment and positions as

$V(\textbf{r}) = \frac{\textbf{p}\cdot(\textbf{r}-\textbf{r}^\prime)}{4\pi\epsilon_0\vert\textbf{r}-\textbf{r}^\prime\vert^3},$

where $\textbf{r}$ and $\textbf{r}^\prime$ are the observation and dipole positions, respectively.

The electric field components for our simple case can be recovered by using

$\textbf{E} = -\nabla V$ .

Carrying out the differentiation in spherical coordinates, we get

$E_r = \frac{Qb}{\pi\epsilon_0 r^3}\cos\theta,$

$E_\theta = -\frac{Qb}{2\pi\epsilon_0 r_3} \sin\theta$ .

These expressions are exactly those used in the previous discussion of dipoles. We see that under many circumstances, the use of potentials can simplify the derivation of vector field problems.

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Distributed charges

We can extend the simple point charge model to distributed systems of charge if we replace the finite point charge with a differential quantity of charge, hence (here, we use $Φ$ to refer to potential, to avoid confusion with the differential volume element dV)

$d\Phi = \frac{dQ}{4\pi\epsilon_0 \vert\textbf{r} -\textbf{r}^\prime\vert}$ .

It is important to recognise that $\vert\textbf{r} - \textbf{r}^\prime\vert\,$ is the distance from the differential charge element and the observation point. Primed coordinates are often used for indicating the source point (where the charge is) and unprimed coordinates are used for the observer's point.

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Example

Write out the integral that describes the potential of a spherically symmetric distribution of charge whose density has the form

$\rho = \rho_0 e^{-\alpha r^\prime}$ ,

where $\rho_0\,$ and $\alpha\,$ are constants; $\alpha\,$ is positive.

Answer:

It is important to recognise that the differential unit of charge is

$dQ = \rho\, dV = \rho_0e^{-\alpha r^\prime} r^{\prime 2}\sin\theta^\prime dr\, d\theta\, d\phi$

so that,

$\Phi = \frac{\rho_0}{4\pi\epsilon_0}\int_0^{2\pi}\!\int_0^\pi\! \int_0^\infty \frac{e^{-\alpha r^\prime}}{\vert\textbf{r}-\textbf{r}^\prime\vert }\sin\theta^\prime \, dr^\prime \, d\theta^\prime \, d\phi^\prime$ .