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I was reading George Green's An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism, and I got confused on one step in his derivation of Poisson's Equation. Specifically, how does Green obtain conclude that: $$\delta\left(2\pi a^2\varrho-\frac{2}{3}\pi b^2\varrho\right)=-4\pi\varrho.$$ Here are two pages to provide context; I understand everything except for the equality above.

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2 Answers 2

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Let's first derive the value of $V$ inside the small sphere: $$ V_\text{sphe} = \rho\int\frac{\text{d}x'\text{d}y'\text{d}z'}{r'}, $$ Where the sphere is sufficiently small such that $\rho$ can be considered constant. We can orientate the axes such that $p$ lies on the $z'$ axis. In spherical coordinates, the integral then has the form $$ \begin{align} V_\text{sphe} &= \rho\int_0^a\text{d}r'\int_0^{2\pi}\text{d}\varphi \int_0^{\pi}\frac{r'^2\sin\theta}{\sqrt{r'^2 + b^2 - 2r'b\cos\theta}}\text{d}\theta\\ &=\frac{2\pi}{b}\rho\int_0^a r' \left(\sqrt{r'^2+b^2+2r'b}-\sqrt{r'^2+b^2-2r'b}\right)\text{d}r'\\ &= \frac{2\pi}{b}\rho\int_0^a r'(r'+b - |r'-b|)\text{d}r'\\ &=\frac{4\pi}{b}\rho\left[\int_0^b r'^2\text{d}r' + \int_b^a r'b\,\text{d}r'\right]\\ &= \frac{4\pi}{3}\rho b^2 + 2\pi\rho a^2 - 2\pi\rho b^2 = 2\pi\rho a^2 - \frac{2\pi}{3}\rho b^2. \end{align} $$ Since $$ b^2 = (x-x_l)^2 + (y-y_l)^2 + (z-z_l)^2, $$ we get $$ \begin{align} \frac{\partial b^2}{\partial x} &= 2(x-x_l),\qquad \frac{\partial^2 b^2}{\partial x^2} = 2 = \frac{\partial^2 b^2}{\partial y^2} = \frac{\partial^2 b^2}{\partial z^2} \end{align} $$ so that $$ \delta b^2 = \frac{\partial^2 b^2}{\partial x^2} + \frac{\partial^2 b^2}{\partial y^2} + \frac{\partial^2 b^2}{\partial z^2} = 6 $$ and $\delta a^2 = 0$ since $a$ is a constant. Therefore, $$ \delta V = \delta V_\text{sphe} = -\frac{2\pi}{3}\rho (\delta b^2) + \left(2\pi a^2 - \frac{2\pi}{3} b^2\right)\delta\rho = -4\pi\rho. $$ The term with $\delta\rho$ disappears since $a$ and $b$ are exceedingly small.

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  • $\begingroup$ Minor comment to the answer (v1): Strictly speaking, in the second-last equality, if $\rho$ is not constant, there is also an (exceedingly small) cross-term contribution from applying the Laplacian $\delta\equiv \Delta$, which is a 2nd-order diff. op. However, since the integration in the first part of the answer (v1) assumes that $\rho$ is constant, it seems unnecessary to later worry about a non-constant $\rho$. $\endgroup$
    – Qmechanic
    Nov 11, 2015 at 17:45
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Pulsar has already given a correct answer. In this answer we will use a slightly different (but equivalent) method, and we will keep a watchful eye on possible distributional contributions.

  1. George Green uses Gaussian units where Coulomb's constant $k_e=1$. He is considering a ball with radius $a$ and uniform charge density $\rho_0$, i.e. of total charge $$ Q~=~\frac{4\pi a^3}{3}\rho_0, \qquad \rho~=~\rho_0\theta(a-r) . \tag{1}$$ Here $\theta(a-r)$ is the Heaviside step function.

  2. Now since Coulomb's law has the same $1/r^2$ dependence as Newtonian gravity, we know from Newton's shell theorem that the electric field $\vec{E}$ is radial, and given by $$ E_r~=~\frac{Q}{r^2}\theta(r-a)+\frac{Qr}{a^3} \theta(a-r) , \tag{2}$$ cf. e.g. this Phys.SE post and links therein.

  3. If we wish, we can integrate to find the potential $$\begin{align} \Phi(r)~=~&\int_r^{\infty}\! dr^{\prime}~E_r(r^{\prime})\cr ~\stackrel{(2)}{=}~& \frac{Q}{r}\theta(r-a)+\frac{Q}{2a}\left(3-\frac{r^2}{a^2}\right) \theta(a-r) \cr ~\stackrel{(1)}{=}~&2\pi \rho_0 \left(\frac{2a^3}{r}\theta(r-a)+\left(a^2-\frac{r^2}{3}\right) \theta(a-r)\right).\end{align}\tag{3}$$

  4. Instead of integrating and then applying the Laplacian, we can also differentiate the electric field (2) directly to deduce that $$\begin{align} - \Delta\Phi~=~& \vec{\nabla}\cdot \vec{E}\cr ~=~&\frac{1}{r^2}\frac{\partial}{\partial r} (r^2E_r)\cr ~\stackrel{(2)}{=}~&\frac{3Q}{a^3}\theta(a-r)\cr ~\stackrel{(1)}{=}~&4\pi\rho_0\theta(a-r)\cr~\stackrel{(1)}{=}~&4\pi\rho.\end{align}\tag{4} $$ Eq. (4) is Poisson's equation in Gaussian's units. It can be viewed as a version of George Green's formula listed in the beginning of OP's post. The rhs. of eq. (4) unsurprisingly confirms that the charge density is $\rho_0$ inside the ball, and zero outside.

  5. Finally, note that when differentiating the Heaviside step functions from eq. (2), that the possible Dirac delta distribution terms $\delta(r-a)$ in eq. (4) precisely cancel, i.e. the ball has no surface charge.

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