Q19P - Introduction to Electrodynamics

Question

Calculate$\mathbf{\nabla }\mathbf{\times }\mathit{E}$directly from Eq. 2.8, by the method of Sect. 2.2.2. Refer to Prob. 1.63 if you get stuck.

Step-by-Step Solution

Answer

The required value is$\nabla \times E\left(r\right)=0$

Step 1: Determine the expression for the electric field in the two sphere.

Write the formula for the electric field of the volume charge.

$\mathrm{E}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \frac{\mathrm{\rho }\left({\mathrm{r}}^{\text{'}}\right)}{{\mathrm{r}}^2}\overline{\mathrm{r}}{\mathrm{d}}_{\mathrm{T}\text{'}}$ 

 

Write the curl of the above function as,

$\nabla \times \mathrm{E}\left(\mathrm{r}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int ∆\times \frac{\left(\overline{\mathrm{r}}\right)}{{\mathrm{r}}^2}\mathrm{r}\text{'}{\mathrm{\rho d}}_{\mathrm{T}\text{'}}$                           ….. (1)

Here, ${\epsilon }_0$is the permittivity of the free space and $\rho$ is the charge density.

Step 2: Determine the value of ∆ × ( r ¯ ) r 2 .

Solve for $∆\times \frac{\left(\overline{\mathrm{r}}\right)}{{\mathrm{r}}^2}$ as,

$$\begin{gathered} ∆\times \frac{\left(\overline{r}\right)}{r^2}=\left(\frac{1}{r^2}\frac{\partial }{\partial r}\overline{r}+\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }\overline{\varphi }\right)\times \left(r^n\varphi \right) \\ =\left|\begin{array}{ccc}\overline{r}& \hat{\theta }& \hat{\varphi }\\ \frac{1}{r^2}\frac{\partial }{\partial r}& \frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }& \frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }\\ r^2& 0& 0\end{array}\right| \\ =\hat{r}\left(0-0\right)-\hat{\theta }\left(0-\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }{r}^2\right)+\hat{\varphi }\left(0-\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }{r}^2\right) \\ =0 \end{gathered}$$

 

Substitute 0 for  $\nabla \times \frac{\left(\hat{r}\right)}{r^2}$in the equation $\nabla \times E\left(r\right)=\frac{1}{4p{\epsilon }_0}\int \nabla \times \frac{\left(\hat{r}\right)}{r^2}{r}^{\text{'}}pd\tau$ .

 $$\begin{gathered} \nabla \times E\left(r\right)=\frac{1}{4p{\epsilon }_0}\left(0\right) \\ =0 \end{gathered}$$

 

Therefore, the required value is $\nabla \times E\left(r\right)=0.$