This question is motivated by Section 3.2.3 in Griffiths.

Therein, we are considering the force of attraction between a point charge and an infinite conducting plane. One can calculate the field using the method of images, and from that the force of attraction exerted by the induced charge to be $$\mathbf{F}=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}\widehat{\mathbf{z}}.$$ One one hand, the energy associated with a field is given by $$W=\frac{\epsilon_0}{2}\int\! E^2\ \mathrm{d}\tau.$$ Since the dot product is positive definite, this implies that $W$ must be positive. On the other hand, the work is also given by \begin{align} W&=\int_\infty^a\! (-\mathbf{F})\cdot \mathrm{d}\mathbf{l}\\ &=\frac{1}{4\pi\epsilon_0}\int_\infty^a \frac{q^2}{4z^2}\ \mathrm{d}z\\ &=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{4d}<0 \end{align} Intuitively, it makes sense that the work should be negative: to bring the particle in from infinity under constant acceleration, we must oppose the attractive force of the induced charge. But I am having trouble reconciling this with the equation $W=(\epsilon_0/2)\int\! E^2\ \mathrm{d}\tau.$ I am sure I am missing something, and would greatly appreciate a nudge in the right direction.

  • $\begingroup$ What is tau? explain all symbols used wheneevr you write $\endgroup$
    – FGSUZ
    Commented Aug 7, 2020 at 14:19
  • $\begingroup$ @FGSUZ The first integral is a volume integral, so $\mathrm{d}\tau$ represents an infinitesimal volume element. $\endgroup$ Commented Aug 7, 2020 at 17:45
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Commented Aug 8, 2020 at 21:06

4 Answers 4


To quote from Griffiths (section 2.4.4 in third edition)

$$ W=\frac12\sum_{i=1}^nq_iV(\mathbf r_i) \tag{2.42}$$ $$ W=\frac{\epsilon_0}{2}\int_\text{all space}E^2\,\text d\tau \tag{2.45}$$

Equation $2.45$ clearly implies that the energy of a stationary charge distribution is always positive. On the other hand, Eq. $2.42$ (from which $2.45$ was in fact derived), can be positive or negative... Which equation is correct?

The answer is that both equations are correct, but they pertain to slightly different situations. Equation $2.42$ does not take into account the work necessary to make the point charges in the first place; we started with point charges and simple found the work required to bring them together. This is wise policy, since Eq. $2.45$ indicates that the energy of a pointcharge is in fact infinite.

$$W=\frac{\epsilon_0}{2(4\pi\epsilon_0)^2}\int\left(\frac{q^2}{r^4}\right)(r^2\sin\theta\,\text dr\,\text d\theta\,\text d\phi)=\frac{q^2}{8\pi\epsilon_0}\int_0^\infty\frac1{r^2}\,\text dr=\infty$$

Equation $2.45$ is more complete in the sense that it tells you the total energy stored in the charge configuration, but Eq. $2.42$ is more appropriate when you're dealing with point charges, because we prefer (for good reason!) to leave out that portion of the total energy that is attributable to the fabrication of point charges themselves.

Griffiths then goes on to describe how the breakdown occurs essentially because a continuous charge distribution is not the same thing as a collection of point particles. i.e. the jump from counting discrete charges to a continuous charge distribution is where the difference occurs. There is no charge located at any single point in a continuous distribution.


I think that this misunderstanding is simply due to different contexts/definitions for the equations.

The first equation, properly written as $$W = \frac{\epsilon}{2} \int E^2 d\tau$$ refers to the electrostatic potential energy stored by an electric field $E$ from a continuous charge distribution. This doesn't apply to a system of two distinct point charges!

The second equation, $$W = \int \vec{F}\cdot d\vec{\ell}$$ refers to the work done by the electric field on a point charge moved through the field.

What you must consider is the tradeoff that occurs as you bring the charge. As you say, negative work is done in opposing the attractive force between the two differently-charged particles. Because they are oppositely-charged, the potential energy decreases as you bring them together, and therefore the change in the stored electrostatic potential energy (the first equation) decreases. That decreased energy goes into opposing your act of bringing in the charge, which you had represented as a negative work (in the second equation).

In the end, both ways are consistent. You just have to remember that difference between the two equations. In the first equation, $W$ is merely a potential energy stored by the field, in the second equation, $W$ is the work done (change in energy!) across the motion defined by the particle's trajectory.

  • 1
    $\begingroup$ "The second equation refers to the work done by the electric field on a point charge moved through the field": If $\mathbf{F}$ is the electric force acting on the particle, as OP has defined it, then the work done by the electric field on the particle will be $W = \int \mathbf{F} \cdot d\mathbf{l}$, i.e. with no negative sign. If you include the negative sign, then your equation is now the work done by an exactly opposing force (e.g. an external force to constrain the particle to be continually in equilibrium during the process). $\endgroup$
    – 13509
    Commented Aug 7, 2020 at 15:24
  • 1
    $\begingroup$ Thanks for the catch. I'll adjust it to represent OP's definition of $\vec{F}$. $\endgroup$
    – zh1
    Commented Aug 7, 2020 at 15:31

The $2$ $W$'s are different quantities. $$W_1=\int\! E^2\ d\tau.$$ $$ W_2=\int_\infty^a\! (-\mathbf{F})\cdot \mathrm{d}\mathbf{l}\\$$ $$W_1-W_2=constant$$ This constant is there because there is quite a bit of potential energy that goes unaccounted while calculating $W_2$. Think of 2 balls of charge separated by a distance $d$. $W_2$ only considers the interactions between the 2 balls. While $W_1$also includes all those interactions within either ball, like the self energy of each ball. In a way, $W_1$ is more complete than $W_2$.

What does electrostatic self-energy mean?


System of two charges.

$\text{Work done by you in bringing two charges together}$


$\text{Change in the electric potential energy of the system of two charges}$ >

If the charges are of opposite sign then work is done on you by the electric field and the electric potential energy of the system of two charges decreases.

The energy stored in an electric field , $E$, in a volume $dV$ is $\frac 12 \epsilon_0 E^2 \,dV$ or you could say the the energy density is $\frac 12 \epsilon_0 E^2$.


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