Q39P - Introduction to Electrodynamics

Question

(a) Check the divergence theorem for the function ${\mathit{v}}_{\mathbf{1}}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{ }\hat{\mathbf{r}}$ , using as your volume the sphere of radius R, centred at the origin.

(b) Do the same for  ${\mathit{v}}_{\mathbf{2}}\mathbf{=}\mathbf{\left(}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{\right)}\hat{\mathbf{r}}$. (If the answer surprises you, look back at Prob. 1.16)

Step-by-Step Solution

Answer

(a) The left and right side of the gauss divergence theorem is equal. Thus, the gauss divergence theorem is proved for $v_1=r^2 \hat{\mathrm{r}}$ .

 

(b) The left and right side of the gauss divergence theorem is not equal. Thus, the gauss divergence theorem is not proved for $v_2=\left(\frac{1}{{\mathrm{r}}^2}\right)\hat{\mathrm{r}}$ .

Step 1: Verify the divergence theorem for function v 1   ( r ) in part (a)

The divergence of function ${\mathit{v}}_{\mathbf{1}\mathbf{ }}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$  is defined as $\mathbf{\nabla }\mathbf{.}{\mathit{v}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}{\mathit{r}}^{\mathbf{2}}{\mathit{v}}_{\mathbf{r}}\mathbf{\right)}$.The function $v_1=r^2 \hat{r}$. Now, find the divergence of  $v_1=r^2 \hat{r}$as: 

$$\begin{gathered} \nabla .v_1=\frac{1}{{\mathrm{r}}^2}\frac{\partial }{\partial \mathrm{r}}(r^2 \hat{r}) \\ =\frac{1}{r^2}4r^3 \\ =4r \end{gathered}$$

The volume integral is taken over the surface of radius R. Now, compute the left part of gauss divergence theorem as:

$$\begin{gathered} \int \left(\nabla .v_1\right)dv=\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }} \underset{0}{\overset{R}{\int }}\left(4r\right)r^2 \sin \theta dr d\theta d\varphi \\ =4\left(\frac{R^2}{4}\right)\underset{0}{\overset{\pi }{\int }}\sin \theta d\theta \underset{0}{\overset{2\pi }{\int }}d\varphi \\ =4\pi R^2 \end{gathered}$$

Taking the area integration over the surface area of sphere of radius R, to calculate the right side of Gauss divergence theorem, the differential area becomes, $da=r^2\sin \theta d \theta d \varphi$ .

 

The right side of Gauss divergence theorem is computed as follows:

$\begin{array}{rcl}\int v_1.da& =& \int (r^2 \sin \theta dr d\theta d\varphi )r^2\\ & =& \underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }} r^4 \sin \theta d\theta d\varphi \\ & =& 4\pi R^2\end{array}$

Thus, the left and right side of gauss divergence theorem are equal. Hence, the theorem is verified for the function in part (a).

Step 2: Verify the divergence theorem for function v 2 = ( 1 r 2 ) r ^ in part (b)

The function is given by $v_2=\left(\frac{1}{{\mathrm{r}}^2}\right)\hat{\mathrm{r}}$ . Now find the divergence of  $v_2=\left(\frac{1}{{\mathrm{r}}^2}\right)\hat{\mathrm{r}}$as:

$$\begin{gathered} \nabla .v_1=\frac{1}{{\mathrm{r}}^2}\frac{\partial }{\partial \mathrm{r}}{r}^2\left(\frac{1}{r^2}\hat{r}\right) \\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(1\right) \\ =0 \end{gathered}$$

The volume integral is taken over the surface of radius R. As$\nabla .v_2=0$  , then the left part of gauss divergence theorem is  computed as:$\int (\nabla -v_1)dv=0$

 Taking the area integration over the surface area of sphere of radius R, to calculate the right side of Gauss divergence theorem, then differential area becomes $da=r^2\sin \theta d \theta d \varphi$ .

The right side of Gauss divergence theorem is computed as follows:

$\begin{array}{rcl}\int v_1.da& =& \int (r^2 \sin \theta dr d\theta d\varphi )\left(\frac{1}{r^2}\right)\\ & =& \underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\pi }{\int }} \sin \theta d\theta d\varphi \\ & =& 4\pi \end{array}$

Thus, the left and right side of gauss divergence theorem are not obtained to be equal. Hence, the theorem is not verified for the function in part (b).