I recently came across the statement (without further explanation) that the net charge in a two dimensional system has to be zero. Obviously in two dimensions, the electric field $\vec{E}$ due to a point charge is inversely proportional to the distance to the charge: $$\vec{E} \propto \frac{\hat{r}}{r}. $$ If we compute the negative line integral of $\vec{E}$ in the radial direction from some $r' = r_0$ to $r' = r$, we get the electric potential $$V(r) = -\int_{r_0}^{r}\vec{E} \cdot d\hat{r} \propto -\ln(r/r_0).$$

Obviously this diverges for $r \to \infty$, but I'm not sure that itself is a problem.

My feeling is that the problem is that a nonzero net charge would lead to infinite electrostatic potential energy, but I'm not sure how to prove that.

Another idea was to consider the energy density $|\vec{E}|^2 \propto 1/r^2$. This would give a total energy proportional to $$\int_{0}^{\infty}\frac{1}{r}dr$$ which diverges. The problem is of course that the corresponding integral in three dimensions, $$\int_{0}^{\infty}\frac{1}{r^2}dr$$ also diverges, so it doesn't seem like a very good argument.

However, we can note that the integral in two dimensions diverges at $\infty$, whereas the integral in three dimensions diverges at $0$.

Any input on this?

EDIT: Here is a translation of the text where it is mentioned. For context: It has just been shown that for an infinite straight line charge with charge density $\lambda$ the potential difference $V(a)-V(b)$ is given by $$V(a)-V(b) = \frac{\lambda}{2\pi\epsilon_0}\ln(b/a).$$

Then comes the following comment: An infinite straight line charge is often called a two dimensional point charge. The electric potential due to the two dimensional point charge at $\vec{r}'$ is usually defined as $$V(\vec{r}) = \frac{\lambda}{2\pi\epsilon_0}\ln(1/|\vec{r}-\vec{r}'|),$$ where $\vec{r} = (x,y), \vec{r}'=(x',y').$ It is seen that the dimension of the argument of the logarithm function is inverse length, which might seem strange.The explanation is that in two dimensional problems the total charge has to be zero. This means that the logarithm functions [due to different point charges] of the electric potential can always be combined in such a way that the argument is dimensionless, such as in $\ln(b/a).$

  • 1
    $\begingroup$ can you provide a source for this statement? Some context would be useful $\endgroup$ Commented Nov 10, 2016 at 23:53
  • $\begingroup$ @BySymmetry Sorry for my late reply. It was mentioned in a set of lecture notes (not in English). $\endgroup$ Commented Nov 11, 2016 at 19:13
  • 2
    $\begingroup$ This is a really strange statement. There's nothing dimensionally wrong with the potential of a line of charge. The final expression just looks wrong because the lecture notes set $b = 1$, while they really meant $b = 1 \text{ meter}$. $\endgroup$
    – knzhou
    Commented Nov 11, 2016 at 19:37
  • 1
    $\begingroup$ Perhaps it is implicitly assumed that the space manifold has no boundary? It's then just Gauss's law in integrable form: The total charge $Q$ is proportional to the electric flux $\Phi_E$ out of the total volume. But if there is no boundary, then $Q \propto \Phi_E=0$. This of course works in any dimensions, not just 2. $\endgroup$
    – Qmechanic
    Commented Nov 11, 2016 at 19:39
  • 1
    $\begingroup$ I agree with @knzhou, it is an odd statement and I don't think your lecturer's argument follows. To make the expression dimensionally consistent you do need to introduce an arbitrary length scale, but because it is inside the logarithm changing this length scale amounts to shifting the potential by a constant, which we are always free to do. $\endgroup$ Commented Nov 11, 2016 at 19:54

1 Answer 1


The following cannot be considered a real answer, but I'm trying to figure what your lecturer meant and to give you some insights accordingly.

The potential difference between points $A$ and $B$ is defined as

$$V_A-V_B=-\int_A^B \vec E \cdot d\vec r$$

If you compute this quantity for a point charge, you obtain

$$V(r_a)-V(r_b)=kq \left(\frac 1{r_a}-\frac 1{r_b}\right)$$

From which we deduce that

$$V(r) = \frac{kq}{r}+C$$

Now, we usually require that $V(\infty)=0$, from which we obtain $C=0$. This is not the only possible choice. We could also require $V(1 \text{m})=0$, which would imply

$$C=-\frac{kq}{1 \text{m}}$$

Now let's consider the charged infinite line. The potential difference is

$$V(r_a)-V(r_b)=\frac{\lambda}{2 \pi \epsilon_0} \ln\left(\frac{r_b}{r_a}\right)$$

from which we deduce that

$$V(r) =- \frac{\lambda}{2 \pi \epsilon_0} \ln\left(\frac r {r_0}\right)+C$$

where $r_0$ is an arbitrary constant with the same dimensions of $r$, which was included to make the argument of the logarithm dimensionless. Now you see that it is not possible to have $V(\infty)=0$ like for a point charge. It is much more natural to impose $V(r_0)=0$, which implies $C=0$. Maybe what your lecturer tried to tell you had something to do with this...?

However, I don't see how the statement that "in two dimensional problems the total charge has to be zero" could be made meaningful.

  • $\begingroup$ Thanks! Yes, I know how the potential works. My question is rather about the cryptic statement about zero total charge in 2D. $\endgroup$ Commented Nov 11, 2016 at 20:59
  • $\begingroup$ @ÉtienneBézout I'm sorry, I tried to figure what that could mean but it just seems like a meaningless statement to me. I thought it could have had something to do with the definition of the potential function in this case, but apparently this is not the case. $\endgroup$
    – valerio
    Commented Nov 11, 2016 at 21:04
  • $\begingroup$ No need to be sorry, I just wanted to clarify. Yes, it's very strange. $\endgroup$ Commented Nov 11, 2016 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.