Electrostatic work and energy

In which we show that an electric field $\myv E(\myv r)$ has an effective energy density proportional to $|E|^2$.

Work

In previous physics courses, you found that work is the product of force $\times$ displacement. Well, actually there are some bits in there about the vector directions of the forces and displacements. To be a bit more precise $$W = - \int_{\myv a}^{\myv b} \myv F \cdot d\myv{l}.$$

The force the cyclist exerts to move his bike uphill is equal and opposite the external force $\myv F$ that he's working against.

The force acting on a particle carrying charge $Q$ in an external electric field $\myv E$ is $\myv F = Q \myv E$, and the work needed to push against this force is.. $$W= -Q \int_{\myv a}^{\myv b} \myv E \cdot d \myv l=Q[V(\myv b)-V(\myv a)].$$

$\Rightarrow$  The potential difference between points $\myv a$ and $\myv b$ is the work per unit charge needed to move a charged body from $\myv a$ to $\myv b$.

If $\myv a$ is our standard reference point at infinity, then "the" electric potential $V(\myv r)$ is the work per unit charge needed to bring a charge from infinity to $\myv r.$

Energy of a charge configuration

Imagine assembling an arrangement of charge by bringing in point charges from our reference point (at infinity) one charge at a time.

No work is done to bring the first charge $q_1$ in to place, since there are no electric fields from other charges. (It doesn't take any work for the charge to move in its own field.)

To move $q_2$ into place, we have to work against the field of the first, doing an amount of work: $$W_2 = q_2 V_1(\myv r_2) = \frac{1}{4\pi \epsilon_0}q_2 \frac{q_1}{\rr_{12}}.$$

Moving the third into place involves moving $q_3$ in the field of the two already-there charges.

The total work to bring in 5 charges is... $$\begineq W = \frac{1}{4\pi\epsilon_0} &(&\\ & & \frac{q_2q_1}{\rr_{12}} + \\ & &\frac{q_3q_1}{\rr_{13}}+\frac{q_3q_2}{\rr_{23}}+\\ & &\frac{q_4q_1}{\rr_{14}}+\frac{q_4q_2}{\rr_{24}}+\frac{q_4q_3}{\rr_{34}}+\\ & &\frac{q_5q_1}{\rr_{15}}+\frac{q_5q_2}{\rr_{25}}+\frac{q_5q_3}{\rr_{35}}+\frac{q_5q_4}{\rr_{45}} \\ & ) &.\endeq$$

Example: Problem 2.31

1. The three charges shown are situated on a square. How much work does it take to bring another charge $+q$ from infinity?
2. How much work did it take to assemble all four charges?
3. If each charge were doubled, how much work would it take to assemble the four charges?

Squaring the triangle...

The total work to bring in 5 charges was... $$\begineq W = \frac{1}{4\pi\epsilon_0} &(&\\ & & \frac{q_2q_1}{\rr_{12}} + \\ & &\frac{q_3q_1}{\rr_{13}}+\frac{q_3q_2}{\rr_{23}}+\\ & &\frac{q_4q_1}{\rr_{14}}+\frac{q_4q_2}{\rr_{24}}+\frac{q_4q_3}{\rr_{34}}+\\ & &\frac{q_5q_1}{\rr_{15}}+\frac{q_5q_2}{\rr_{25}}+\frac{q_5q_3}{\rr_{35}}+\frac{q_5q_4}{\rr_{45}} \\ & ) &.\endeq$$

Let's write a sum where we "square up" the triangular tendencies of the sum above $$W=\frac{1}{8 \pi \epsilon_0} \sum_{i=1}^n \sum_{j=1}^{n; i\ne j} \frac{q_i q_j}{\rr_{ij}}.$$

This sum includes the original term with $\rr_{14}$ as well as a new term with $\rr_{41}$ in it--but of the same magnitude, and so we've just divided by 2 to account for the double counting. Re-writing this as... $$\begineq W &=& \frac{1}{2} \sum_{i=1}^n q_i \sum_{j=1}^{n; i\neq j} \frac{1}{4\pi \epsilon_0} \frac{q_j}{\rr_{ij}} \\ &=&\frac{1}{2}\sum_{i=1}^n q_i V(\myv r_i),\endeq$$

where $V(\myv r_i)$ is the potential at $\myv r_i$ due to all the other charges in the system.

If all the charged bodies were removed as far as possible from each other, this is the energy that would be liberated.

Energy of a continuous charge distribution

For point changes, we just had... $$W=\frac{1}{2}\sum_{i=1}^n q_i V(\myv r_i).$$

In this expression, $V(\myv r_i)$ is the potential due to all the charges except $q_i$.

But if we consider a charge density in the limit as $dq=\rho(\myv r)d\tau$ becomes vanishingly small...
Then the potential due to all the charges except $dq$ is vanishingly different from just the potential due to *all* the charges at the position of $dq$.

With $V(\myv r)$ now meaning just "the" potential at $\myv r$ due to *all* the charges, we can write the sum as an integral: $$W=\frac{1}{2}\int \rho(\myv r) V(\myv r)\, d \tau.$$

This integral must at least encompass all the places where the charge density is non-zero. It wouldn't hurt to simply extend it over all space, for if there are places where $\rho$ vanishes, those don't contribute anyway to the integral.

There's a way of writing this in terms of the electric field instead of $\rho$ and $V$.

Using Poisson's equation, $\grad^2 V=-\rho/\epsilon_0$ the work expression becomes $$W=-\frac{\epsilon_0}{2}\int_{\infty} V \grad^2 V d \tau.$$

The integrand looks a bit like one term in the derivative of a product. It is.

Using product rule (5) (inside your front cover): $$\myv \grad \cdot(V \myv \grad V) = (\myv \grad V)\cdot(\myv \grad V)+V (\grad^2 V) .$$

Re-arranging this and substituting into the expression for work: $$W=-\frac{\epsilon_0}{2}\int_{\infty } \myv \grad\cdot (V \myv \grad V)\,d\tau + \frac{\epsilon_0}{2} \int_{\infty} \myv \grad V\cdot\myv \grad V\, d \tau.$$

Using the divergence theorem, the first volume integral can be re-written as a surface integral: $$\int_{\infty } \myv \grad\cdot(V \myv\grad V)\,d\tau = \oint_{\infty {\cal S}} V \myv \grad V \cdot d \myv a = 0.$$

It vanishes because $V=0$ on the surface (at infinity) where the integral is being evaluated. This leaves just one term in the work expression. Since $\myv E=- \myv \grad V$:

$$W = \frac{\epsilon_0}{2} \int_{\infty} |E^2| d \tau.$$

Therefore the total amount of energy required to assemble any charge distribution is always a positive number (see problem 2.31)...right?

Find the energy of configuration of a uniformly charged solid sphere of radius $R$ using the $E^2$ integral.
Use $E_{r\, \text{ inside}} = \frac{1}{4\pi\epsilon_0} \frac{q}{R^3} r\uv{r}$ and $E_{r\,\text{ outside}} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} \uv{r}$.
Answer: $W = \frac{1}{4\pi \epsilon_0}\frac{3}{5}\frac{q^2}{R}.$