📂Electrodynamics

Work and Energy in Electrostatics

Moving a Charge¹

The following equation holds between potential and electric field.

$$-\int_\mathbf{a} ^\mathbf{b} \mathbf{E} \cdot d\mathbf{l} = \int_\mathbf{a} ^ \mathbf{b} \left( \nabla V \right) \cdot d\mathbf{l} = V(\mathbf{b}) - V(\mathbf{a})$$

Thus, if there is a fixed source charge distribution, and we move a test charge $Q$ from point $\mathbf{a}$ to point $\mathbf{b}$, the work done is calculated as follows.

$$W=\int_{\mathbf{a}}^\mathbf{b} \mathbf{F} \cdot d\mathbf{l} = -Q\int_\mathbf{a}^\mathbf{b} \mathbf{E} \cdot d\mathbf{l} =Q[V(\mathbf{b})-V(\mathbf{a})]$$

Dividing the above equation by $Q$ yields this. It means that the potential difference between $\mathbf{a}$ and $\mathbf{b}$ is equal to the work done to move the charge $Q$ from $\mathbf{a}$ to $\mathbf{b}$.

$$V(\mathbf{b})-V(\mathbf{a})=\dfrac{W}{Q}$$

The work done to move charge $Q$ from a very distant place to location $\mathbf{r}$ is

$$W=Q[V(\mathbf{r}) - V( \infty ) ]=QV(\mathbf{r})$$

Energy of a Point Charge Distribution

Let’s assume a charge $q_{1}$ was moved from a very distant place into empty space. Since it was an empty space, there is no electric field, and thus, no work is done.

Next, let’s say we move charge $q_2$ from a distant place to $\mathbf{r}_2$. Now that $q_{1}$ exists, there’s an electric field, and work is done to move $q_2$. Let’s call the value of the potential created by $q_{1}$ at $\mathbf{r}_2$ as $V_{1}(\mathbf{r}_2)$, and the distance between $q_{1}$ and $q_2$ after $q_2$ has been moved as $\cR_{12}$. Then, according to the formula derived above,

$$W_2=q_2V_{1}(\mathbf{r}_2)=q_2\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}}{\cR_{12}}$$

Similarly, let’s say we move charge $q_{3}$ from a distant place to $\mathbf{r}_{3}$. Since it is influenced by the electric field created by $q_{1}, q_2$,

$$W_{3}=q_{3}\left[ V_{1}(\mathbf{r}_{3})+V_{2}(\mathbf{r}_{3}) \right] =q_{3}\dfrac{1}{4\pi\epsilon_{0}} \left( \dfrac{q_{1}}{\cR_{13}}+\dfrac{q_2}{\cR_{23} }\right)$$

Likewise, if we bring in charge $q_{4}$,

$$W_{4}=q_{4}\left[ V_{1}(\mathbf{r}_{4}) +V_2(\mathbf{r}_{4}) +V_{3}(\mathbf{4}) \right]=q_{4}\dfrac{1}{4\pi\epsilon_{0}} \left( \dfrac{q_{1}}{\cR_{14}}+\dfrac{q_2}{\cR_{24}} +\dfrac{q_{3}}{\cR_{34}}\right)$$

Thus, the total work done in bringing the first 4 charges together is

$$W=\dfrac{1}{4\pi\epsilon_{0}}\left(\dfrac{q_{1}q_2}{\cR_{12}} +\dfrac{q_{1}q_{3}}{\cR_{13}}+\dfrac{q_{1}q_{4}}{\cR_{14}}+\dfrac{q_2q_{3}}{\cR_{23}}+\dfrac{q_2q_{4}}{\cR_{24}}+\dfrac{q_{3}q_{4}}{\cR_{34}} \right)$$

Generalizing it for $n$ charges,

$$W=\dfrac{1}{2}\dfrac{1}{4\pi\epsilon_{0}}\sum \limits_{i}^n \sum \limits_{j \ne i } ^n \dfrac{q_{i}q_{j}}{\cR_{ij}}$$

Since the case with $i=2$, $j=3$ and the case with $i=3$, $j=2$ are the same, we divide the total cases by $2$. Summarizing,

$$W=\dfrac{1}{2}\sum \limits_{i}^n q_{i} \left( \sum \limits_{j \ne i} ^n \dfrac{1}{4 \pi \epsilon_{0}}\dfrac{q_{j}}{\cR_{ij}} \right)$$

At this point, the terms inside the parentheses are the potentials created by <$q_{j}$s at the location where $q_{i}$ is present and $\mathbf{r}_{i}$ is present with charge $j\ne i$,

$$W=\dfrac{1}{2}\sum \limits_{i}^n q_{i}V(\mathbf{r}_{i})$$

This represents the energy stored in the gathered point charges.

Energy of a Continuous Charge Distribution

If we represent the formula derived above for point charges in terms of line charge density, surface charge density, and volume charge density, respectively,

$$\begin{align*} W =&\ \dfrac{1}{2} \int \lambda V dl W =&\ \dfrac{1}{2} \int \sigma V da W =&\ \dfrac{1}{2} \int \rho V d\tau \end{align*}$$

In the formula for volume charge, if we use Gauss’s law $\nabla \cdot \mathbf{E} = \dfrac{\rho}{\epsilon_{0}}$ to replace $\rho$ with $\mathbf{E}$,

$$W=\dfrac{\epsilon_{0}}{2} \int (\nabla \cdot \mathbf{E}) V d\tau$$

This formula can be further transformed using integration by parts as follows.

$$W=\dfrac{\epsilon_{0}}{2} \left[ -\int \mathbf{E} \cdot (\nabla V)d\tau + \oint V\mathbf{E} \cdot d\mathbf{a} \right]$$

Since $\nabla V=-\mathbf{E}$,

$$W= \dfrac{\epsilon_{0}}{2} \left[ \int _\mathcal{V} E^2 d\tau + \oint_\mathcal{S}V\mathbf{E}\cdot d\mathbf{a} \right]$$

In this case, the integration domain can be any region as long as it encompasses the charge. Since the charge density in a region without charge is $\rho=0$, as long as all charges are included, the integration domain can be made infinitely large without affecting the result. Examining the first term reveals that as the integration domain expands, the integral value continues to increase since the integrand is positive. Because the total value is fixed at $W$, the magnitude of the first term must increase while the value of the surface integral, the second term, continues to decrease. Therefore, integrating over the entire space,

$$W=\dfrac{\epsilon_{0}}{2} \int_{\mathrm{total\ space}} E^2 d\tau$$