Let $\mathbf{F}=\mathbf{F}(x,y,z)$ be an electric force field $$\mathbf{F}=\mathbf{F}(x,y,z)=\frac{\varepsilon qQ}{x^{2}+y^{2}+z^{2}}\mathbf{u}$$

where $\varepsilon$ is a constant that depends on the units used; $q$ and $Q$ are the charges of two objects, one located at the point $(0,0,0)$ and the other at a point $(x,y,z)$; and $\mathbf{u}$ is the unit vector in the direction from $(0,0,0)$ to $(x,y,z)$. Let $S$ be a closed surface with positive orientation that encloses a solid $E$ in space.

Show that the following statements are true under the assumptions of the Divergence Theorem:

Solution The unit vector $\mathbf{u}=\dfrac{x\mathbf{i}+y\mathbf{j }+z\mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}$, so that $$\mathbf{F}=\mathbf{F}(x,y,z)=\varepsilon qQ\dfrac{x\mathbf{i}+y\mathbf{j}+z \mathbf{k}}{(x^{2}+y^{2}+z^{2})^{3/2}}$$

The electric force field $\mathbf{F}$ is continuous everywhere except at the point $(0,0,0)$.

(a) The surface $S$ that forms the boundary of the solid $E$ satisfies the assumptions of the Divergence Theorem. We find ${\rm{div}} \mathbf{F}$. $$\begin{eqnarray*} {\rm{div}}\mathbf{F}&= &\varepsilon qQ\left\{\dfrac{\partial }{\partial x}\left[ \dfrac{x}{ (x^{2}+y^{2}+z^{2})^{3/2}}\right] +\dfrac{\partial }{\partial y}\left[ \dfrac{y }{(x^{2}+y^{2}+z^{2})^{3/2}}\right] +\dfrac{\partial }{\partial z}\left[ \dfrac{z}{(x^{2}+y^{2}+z^{2})^{3/2}}\right] \right\} \\ &=&\varepsilon qQ\bigg\{\dfrac{(x^{2}+y^{2}+z^{2})^{3/2}-3x^{2} \sqrt{x^{2}+y^{2}+z^{2}}}{(x^{2}+y^{2}+z^{2})^{3}}+\dfrac{ (x^{2}+y^{2}+z^{2})^{3/2}-3y^{2}\sqrt{x^{2}+y^{2}+z^{2}}}{ (x^{2}+y^{2}+z^{2})^{3}} \\ &&+\,\dfrac{(x^{2}+y^{2}+z^{2})^{3/2}-3z^{2}\sqrt{x^{2}+y^{2}+z^{2}}}{ (x^{2}+y^{2}+z^{2})^{3}} \bigg\} \\ &=&\varepsilon qQ\dfrac{3(x^{2}+y^{2}+z^{2})^{3/2}-3\sqrt{x^{2}+y^{2}+z^{2}}(x^{2}+y^{2}+z^{2})}{(x^{2}+y^{2}+z^{2})^{3}}=0 \end{eqnarray*}$$

Therefore, by the Divergence Theorem, $$\iint\limits_{S}\mathbf{F}\,{\cdot}\, \mathbf{n}\,dS=\iiint\limits_E {\rm{div}} \mathbf{F}\,dV=0$$

Figure 70

(b) Since the interior of $S$ contains the origin, $\mathbf{F}$ is not continuous at the origin, so we cannot use the Divergence Theorem. We use the following argument to prove (b). Let $E$ be the closed solid enclosed by two separate surfaces: the surface $S$ and a sphere $S_{a}$ of radius $a$, with its center at $(0,0,0)$, as shown in Figure 70. The outer surface is $S$ and the inner surface is $S_{a}$. Now $\mathbf{F}$ is continuous throughout $E$, so the Divergence Theorem can be used. $$\iiint\limits_E {\rm{div}}\mathbf{F}\,dV=\iint\limits_{S}\mathbf{F}\,{\cdot}\, \mathbf{n}\,dS+\iint\limits_{S_{a}}\mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS$$

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From (a), we know that $\iiint\limits_{E}{\rm{div}}\mathbf{F}\,dV=0$. So, $$\iint\limits_{S}\mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=-\iint\limits_{S_{a}}\mathbf{F }\,{\bf\cdot}\, \mathbf{n}\,dS$$

On the inner surface $S_{a}$, a sphere of radius $a$, the outer unit normal is $$\mathbf{n}=-\dfrac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{a}=-\dfrac{x\mathbf{i}+ y\mathbf{j}+z\mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}$$

Then $$\begin{eqnarray*} \mathbf{F}\,{\bf\cdot}\, \mathbf{n\,} &=&\dfrac{\varepsilon qQ}{x^{2}+y^{2}+z^{2}} \mathbf{u}\,{\bf\cdot}\, \left( -\dfrac{x\mathbf{i}+y\mathbf{j}+z\mathbf{k}}{\sqrt{ x^{2}+y^{2}+z^{2}}}\right)\\ &=&-\dfrac{\varepsilon qQ}{x^{2}+y^{2}+z^{2}}\dfrac{x \mathbf{i}+y\mathbf{j}+z\mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}\,{\bf\cdot}\, \dfrac{x \mathbf{i}+y\mathbf{j}+z\mathbf{k}}{\sqrt{x^{2}+y^{2}+z^{2}}} \\ &=&-\varepsilon qQ\dfrac{x^{2}+y^{2}+z^{2}}{a^{4}}=-\dfrac{\varepsilon qQ}{ a^{2}} \end{eqnarray*}$$

So,