Total flux through a spherical surface enclosing an off-centered source of uniformly distributed lines

Consider a point source $Q$ in space from which $N$ number of lines emanate uniformly in all directions. This source is enclosed in a spherical shell of radius $R$ and is placed a distance $d$ from the center of the sphere. Figure 1 (a) illustrates this arrangement.

Figure 1 Let the symbol $S$ represent the surface of the spherical shell. The flux through an infinitesimal area of the surface of the sphere is given by $$\vec{\rho}(\vec{r}) \cdot \vec{dA}$$ where $\vec{\rho}(\vec{r})$ is the density of line at a point $\vec{r}$ on $S$ and $\vec{dA}$ is the infinitesimal area vector at $\vec{r}$ on $S$. It can be shown that the total outward flux of lines through the sphere is $$\oint_{S} \vec{\rho}(\vec{r}) \cdot \vec{dA} = N$$ Figure 1(b) illustrates the geometry of the problem, manner of placement of the coordinate system and various symbols used to represent variables in the problem. Note that the coordinate system has been chosen such that both the source and the center of the sphere are located on the $y$-axis. This is done only to simplify the problem mathematically; however, the choice of coordinate system does not affect validity of the statement about the net flux through the spherical surface.

Proof: Referring to figure 1(b), we note that the density of lines at a point $\vec{r'}$ with respect to the source $Q$ is given as $$\vec{\rho}(\vec{r'}) = \frac{N}{4\pi r'^{2}} \hat{r'} .$$ Given the symmetry in the geometry of the problem, spherical polar coordinate system would be a suitable to express this problem in. The infinitesimal area vector indicated in figure 1(b) as $dA$ can be expressed in terms of spherical polar coordinates $(r,\theta,\phi)$ as $$\vec{dA} = R^{2} \sin \theta \, d\theta \, d\phi \, \hat{r}$$ Flux of lines passing through the infinitesimal area $\vec{dA}$ is given by $$\vec{\rho}(\vec{r'}) \cdot \vec{dA} = \frac{N}{4\pi r'^{2}} \hat{r'} \cdot R^{2} \sin \theta \, d\theta \, d\phi \, \hat{r}$$ Net outward flux through the spherical surface of radius $R$ is obtained by integrating the above equation over the entire surface of the sphere. Therefore, $$\begin{eqnarray} \oint \vec{\rho}(\vec{r'}) \cdot \vec{dA} & = & \oint_{S} \frac{N}{4\pi r'^{2}} \hat{r'} \cdot R^{2} \sin \theta \, d\theta \, d\phi \, \hat{r} \\ & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{1}{r'^{2}}\sin \theta \, d\theta \, d\phi \, \hat{r'} \cdot \hat{r} \,\,\,\,\,\,\,......(1) \end{eqnarray}$$ Remainder of the problem consists of solving the above integral. If we represent the angle between the unit vectors $\hat{r'}$ and $\hat{r}$ by $\eta$, then $\hat{r'} \cdot \hat{r} = \cos \eta$. Substituting this in (1), we get $$\begin{eqnarray} \oint \vec{\rho}(\vec{r'}) \cdot \vec{dA} & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{1}{r'^{2}}\sin \theta \, d\theta \, d\phi \, \cos \eta \\ & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{\cos \eta}{r'^{2}}\sin \theta \, d\theta \, d\phi \,\,\,\,\,\,\,......(2) \end{eqnarray}$$ In triangle OPA, $$\begin{eqnarray} d^{2} & = & R^{2} + r'^{2} - 2Rr'\cos \eta \,\,\,\,\,\,\,......(3) \\ r'^{2} & = & R^{2} + d^{2} - 2Rd\cos \theta \,\,\,\,\,\,\,......(4) \end{eqnarray}$$ Rearranging (3), we get $$\begin{equation} \cos \eta = \frac{R^{2} + r'^{2} - d^{2}}{2Rr'} \,\,\,\,\,\,\,......(5) \end{equation}$$ Substituting (5) and then (4) in (2) eliminates $r'$ from the integrand and gives us an integrand that is a function of $r$ only. With these substitutions, we can write $\vec{\rho}$ as a function of $\vec{r}$: $$\begin{eqnarray*} \oint \vec{\rho}(\vec{r}) \cdot \vec{dA} & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{R^{2} + r'^{2} - d^{2}}{2Rr'^{3}}\sin \theta \, d\theta \, d\phi \\ & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{R^{2} - d^{2} + R^{2} + d^{2} - 2Rd\cos \theta}{2R\sqrt{R^{2} + d^{2} - 2Rd\cos \theta}^{3}}\sin \theta \, d\theta \, d\phi \\ & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{2R^{2} - 2Rd\cos \theta}{2R(R^{2} + d^{2} - 2Rd\cos \theta)^{3/2}}\sin \theta \, d\theta \, d\phi \\ & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{R - d\cos \theta}{(-R^{2} +2R^{2} + d^{2} - 2Rd\cos \theta)^{3/2}}\sin \theta \, d\theta \, d\phi \\ & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{R - d\cos \theta}{ [2R(R - d\cos \theta) + d^{2} - R^{2}]^{3/2}}\sin \theta \, d\theta \, d\phi \\ & = & \frac{N R^{2}}{4\pi} \oint_{S} \frac{R - d\cos \theta}{ [2R(R - d\cos \theta) - (R^{2} - d^{2})]^{3/2}}\sin \theta \, d\theta \, d\phi \end{eqnarray*}$$ Taking $a = 2R$ and $b = R^{2} - d^{2}$, and substituting these in the above equation yields $$\oint \vec{\rho}(\vec{r}) \cdot \vec{dA} = \frac{N R^{2}}{4\pi} \oint_{S} \frac{R - d\cos \theta}{ [a(R - d\cos \theta) - b]^{3/2}}\sin \theta \, d\theta \, d\phi$$ The right-hand side of the above equation contains two variables $(\theta, \phi)$ over which we need to integrate to arrive at the solution. Coordinate $\theta$ is to be integrated over the range $[0, \pi]$ and and coordinate $\phi$ over the range $(0, 2\pi)$: $$\begin{equation} \oint \vec{\rho}(\vec{r}) \cdot \vec{dA} = \frac{N R^{2}}{4\pi} \int_{0}^{\pi} \frac{R - d\cos \theta}{ [2R(R - d\cos \theta) - (R^{2} - d^{2})]^{3/2}}\sin \theta \, d\theta \, \int_{0}^{2\pi} \, d\phi \,\,\,\,\,\,\,......(6) \end{equation}$$ Coordinate $\phi$ can be readily integrated the given range to give $2\pi$. The integrand corresponding to the coordinate $\theta$ can be solved by substitution method. Define a new variable $$\begin{equation} y = 2R(R - d\cos \theta) - (R^{2} - d^{2}) \,\,\,\,\,\,\,......(7) \end{equation}$$ Then, $dy = 2Rd \sin \theta \, d\theta.$

According to equation (7), as $\theta \rightarrow 0$, $y \rightarrow R^{2} - 2Rd + d^{2} = (R-d)^{2}$, and $\theta \rightarrow \pi$, $y \rightarrow R^{2} + 2Rd + d^{2} = (R+d)^{2}$. Making these substitutions in (6), we get $$\begin{eqnarray*} \oint \vec{\rho}(\vec{r}) \cdot \vec{dA} & = & \frac{N R^{2}}{4\pi} \int_{0}^{\pi} \frac{R - d\cos \theta}{ [2R(R - d\cos \theta) - (R^{2} - d^{2})]^{3/2}}\sin \theta \, d\theta \, \int_{0}^{2\pi} \, d\phi \\ & = & \frac{N R^{2}}{4\pi} \frac{1}{4R^{2}d} \int_{0}^{\pi} \frac{2R(R - d\cos \theta)}{ [2R(R - d\cos \theta) - (R^{2} - d^{2})]^{3/2}} 2Rd \, \sin \theta \, d\theta \, (2\pi) \\ & = & \frac{N}{8d} \int_{(R-d)^{2}}^{(R+d)^{2}} \frac{y + (R^{2} - d^{2})}{ y^{3/2}} dy \\ & = & \frac{N}{8d} \left( \int_{(R-d)^{2}}^{(R+d)^{2}} \frac{y}{ y^{3/2}} dy + (R^{2} - d^{2}) \, \int_{(R-d)^{2}}^{(R+d)^{2}} \frac{1}{ y^{3/2}} dy \right)\\ & = & \frac{N}{8d} \left( \int_{(R-d)^{2}}^{(R+d)^{2}} y^{-1/2} dy + (R^{2} - d^{2}) \, \int_{(R-d)^{2}}^{(R+d)^{2}} y^{-3/2} dy \right)\\ & = & \frac{N}{8d} \left( 2[y^{1/2}]_{(R-d)^{2}}^{(R+d)^{2}} - 2(R^{2} - d^{2}) \, [y^{-1/2}]_{(R-d)^{2}}^{(R+d)^{2}} \right) \\ & = & \frac{N}{8d} \left[ 2(2d) - 2(R^{2} - d^{2}) \, \frac{-2d}{R^{2} - d^{2}} \right] \\ & = & \frac{N}{8d} \, 8d \\ & = & N \end{eqnarray*}$$ Hence, $$\oint \vec{\rho}(\vec{r}) \cdot \vec{dA} = N.$$

If we replace $N$ in this problem by $Q/\epsilon_{0}$, we get: $$\vec{\rho}(\vec{r}) = \frac{N}{4\pi r^{2}}\hat{r} = \frac{Q}{4\pi \epsilon_{0} r^{2}}\hat{r} = \vec{E}(\vec{r})$$ where $\vec{E}(\vec{r})$ represents electric field strength at a point $\vec{r}$ from a charge $Q$. Further, $\oint \vec{\rho}(\vec{r}) \cdot \vec{dA} = N$ gives: $$\oint \vec{E}(\vec{r}) \cdot \vec{dA} = \frac{Q}{\epsilon_{0}}.$$ This means that the total electric flux through a spherical surface enclosing a charge $Q$ is $Q/\epsilon_{0}$ regardless of position of the charge within the sphere.