Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Electric potential of thin wire by volume integral

3 votes

A well known solution formula to Poisson's equation
$$ \nabla^2 \Phi =\frac{\varrho}{\varepsilon_0}$$
for the electric potential $\Phi$ given the charge density $\varrho$ in electrostatics is (refer e.g. Jackson's electrodynamics)
$$\Phi = \frac{1}{4 \pi \varepsilon_0} \int dV' \frac{\varrho}{|r-r'|} $$
Now, for an infinitely thin wire the charge density in cylindrical coordinates is:
$$ \varrho=\lambda \delta(\varphi) \frac{\delta(\rho)}{\rho} $$
Here $\lambda$ is the line charge density, $\varphi$ the angle and $\rho$ the radius in cylindrical coordinates and $\delta$ the delta-distribution.

Plugging that into the solution formula for the electric potential and doing the radial and angular integrations yields:
$$\Phi = \frac{\lambda}{4 \pi \rho \varepsilon_0} \int_{\mathbb{R}} dz' \frac{1}{\sqrt{1+(\frac{z-z'}{\rho})^2}} $$
The antiderivative of $\sqrt{1+x^2}^{-1}$ is the inverse function of $\sinh(x)$, $\textrm{arsinh}(x)$. The problem now is - $\lim_{x \to \pm \infty}\textrm{arsinh}(x)=\pm \infty$, so I'm not sure of the value of the integral - and thus of the potential for the thin wire. Any help will be appreciated.

asked Nov 7, 2016 in Physics Problems by Sanya (240 points)
reopened Nov 8, 2016 by Einstein

1 Answer

2 votes

The volume integral expression for the electric potential that you show is correct only up to a harmonic function (i.e., a function that satisfies $\nabla^2 f =0$) that depends on the boundary conditions of the problem. This expression for the electric potential given in Jackson is derived from an expression for the electric field, which in turn is derived by applying the Coulomb's law inside a volume V , i.e., by assuming that charges exist only inside the volume V.
$$\mathbf{E}(\mathbf{r}) = \int_V \rho(\mathbf{r'})\frac{\hat{r}-\hat{r}'}{|\mathbf{r}-\mathbf{r'}|^2} dV' = - \nabla \int_V \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|} dV'=-\nabla \Phi(\mathbf{r})$$
Even when this assumption holds true, the electric potential can differ from the expression shown in the question by a constant, because the expression for the electric field remains the same even when a constant is added to the electric potential. However, Jackson chooses this constant as zero, because it assumes that value when the charges exist in a bounded region, i.e., when they do not extend to infinity.
In the more general case i.e., when there are charges outside the volume V or on its surface, the volume integral expression for the electric potential that you show differs from the actual electric potential not just by a constant but by a harmonic function.

In your problem, the charges extend to infinity and the harmonic function is not bounded at many points. This poses certain problems when you try to obtain the electric potential using the volume integral equation. To see why this is so, we need to consider the problem in terms of the Green's function.

The Green's function $G(\mathbf{r},\mathbf{r'})$ for the Poisson's equation satisfies
$$\nabla^2 G(\mathbf{r},\mathbf{r'}) = \delta(\mathbf{r}-\mathbf{r'}) $$

By application of the Green's theorem, the following relation between the electric potential and the Green's function can be shown

$$\Phi(\mathbf{r}) = \int_V G(\mathbf{r},\mathbf{r'}) f(\mathbf{r'}) dV' + \int_S \left[G(\mathbf{r},\mathbf{r'}) \nabla' \Phi(\mathbf{r'})\cdot \hat{n} + \Phi(\mathbf{r'}) \nabla' G(\mathbf{r},\mathbf{r'})\cdot \hat{n} \right] dS $$

where $ f(\mathbf{r'}) = \varrho(\mathbf{r'}) / \varepsilon_0 $, the surface integral (second expression) is taken over the surface $S$ bounding the volume $V$. If the Green's function is assumed to go to zero as $\mathbf{r}$ goes to infinity, the first equation can be solved to show that

$$G(\mathbf{r},\mathbf{r'}) = \frac{1}{4 \pi |\mathbf{r}-\mathbf{r'}|}$$

Plugging this back in the expression for the electric potential, we see that first term becomes the volume integral expression shown in the question. The surface integral term evaluates to a harmonic function. When the charges are bounded by a region, say $M$, we can show that this function becomes zero. To see this, note that we can always choose the volume $V$ to be large enough that the volume of $M$ is much much smaller than $V$. Then $|\mathbf{r}-\mathbf{r'}| \approx |\mathbf{r'}| $. Now assuming that the surface integral evaluates to zero and evaluating the expression for the electric potential for very large $|\mathbf{r'}|$ gives
$$\Phi(\mathbf{r'})= \frac{Q}{4 \pi \varepsilon_0 R } + O\left(\frac{1}{R^2}\right)$$
where $R = |\mathbf{r'}|$. We will now show that plugging this expression back in the surface integral makes it zero, consistent with our assumption. The surface integral can be approximately written as

&\int_S \left[ G \nabla' \Phi \cdot \hat{r}' + \Phi \nabla' G \cdot \hat{r}' \right] R^2 dR\ d\theta\ d\phi \\
&= \int_S \left[ G \frac{ \partial \Phi}{\partial R} + \Phi \frac{\partial G}{\partial R} \right] R^2 dR\ d\theta\ d\phi

Since $G$ is approximately equal to $1/4\pi R$, the above expression reduces to

$$\frac{1}{4\pi} \int_S \left[ R \frac{\partial \Phi}{\partial R}- \Phi\right] dR\ d\theta\ d\phi$$

Substituting the approximate expression for the electric potential that we derived above, we can see that the integrand is of order $1/R^2$ and therefore the integral must go to zero. So this shows that $\Phi$ can be expressed as volume integral.

In your question, if we want to use the expression for the Green's function as $1/4\pi|\mathbf{r}-\mathbf{r'}|$, we have to take into account the non-zero value of the surface integral, which goes to infinity (for all $\mathbf{r}$) if all the charges have to be included in the volume $V$. Therefore, the problem cannot be solved this way. Instead we can use a different Green's function by giving up the assumption that the Green's function goes to zero as $\mathbf{r}$ goes to infinity. For problems with azimuthal symmetry, assuming that the Green's function takes a finite value at some finite non-zero radius gives a Green's function with logarithmic variation in $\rho$. This is the standard Green's function for the Poisson's equation in 2D.

answered Nov 10, 2016 by GP (140 points)
edited Nov 17, 2016 by GP