Question

Show that the electric field E of a point charge [equation (8.11)] is conservative. Write φ in (8.13) in rectangular coordinates, and find $\mathbf{E}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathit{\phi }$using both rectangular coordinates (6.3) and cylindrical coordinates. Verify that your results are equivalent to (8.11).

Short answer

The electric field is conservative.$\nabla \times \mathrm{E}=0$

The electric field is mentioned below.

$$\begin{gathered} \mathrm{E}=q(x\mathrm{i}+y\mathrm{j}+z\mathrm{k})\frac{1}{{\left(x^2+y^2+z^2\right)}^{3/2}} \\ \mathrm{E}=\frac{q\left(s{\mathrm{e}}_s+z{\mathrm{e}}_z\right)}{{\left(s^2+z^2\right)}^{3/2}} \end{gathered}$$

Step-by-step solution

Given Information

The force field is$\nabla \times \mathrm{E}=0$

Definition of conservative force and scalar potential

A force is said to be conservative if $\mathbf{\nabla }\mathbf{\times }\mathit{F}\mathbf{=}\mathbf{0}$.

The scalar potential is independent of the path. The scalar potential is the sum of potential in all the 3 dimensions calculated separately.

The formula for the scalar potential is$\mathit{W}\mathbf{=}\mathbf{\int }\mathit{F}\mathbf{.}\mathit{d}\mathit{r}$

Verify whether the electric field is conservative or not.

The electric field is said to be conservative if$\nabla \times E=0$

$\nabla \times \mathrm{E}=\left|\begin{array}{ccc}i& j& k\\ \partial /\partial x& \partial /\partial y& \partial /\partial z\\ \frac{qx}{{\left(x^2+y^2+z^2\right)}^{3/2}}& \frac{qy}{{\left(x^2+y^2+z^2\right)}^{3/2}}& \frac{qz}{{\left(x^2+y^2+z^2\right)}^{3/2}}\end{array}\right|$

Now solve the determinant.

$\begin{array}{rcl}(\nabla \times \mathrm{E}).\mathrm{i}& =& \left(\frac{-3qyz{\left(x^2+y^2+z^2\right)}^{-3/2}}{{\left(x^2+y^2+z^2\right)}^3}-\frac{-3qyz{\left(x^2+y^2+z^2\right)}^{-3/2}}{{\left(x^2+y^2+z^2\right)}^3}\right)\\ & =& 0\end{array}$

$\begin{array}{rcl}(\nabla \times \mathrm{E})·\mathrm{j}& =& -\left(\frac{-3qxz{\left(x^2+y^2+z^2\right)}^{-3/2}}{{\left(x^2+y^2+z^2\right)}^3}-\frac{-3qxz{\left(x^2+y^2+z^2\right)}^{-3/2}}{{\left(x^2+y^2+z^2\right)}^3}\right)\\ & =& 0\\ (\nabla \times \mathrm{E})·\mathrm{k}& =& \left(\frac{-3qyz{\left(x^2+y^2+z^2\right)}^{-3/2}}{{\left(x^2+y^2+z^2\right)}^3}-\frac{-3qzy{\left(x^2+y^2+z^2\right)}^{-3/2}}{{\left(x^2+y^2+z^2\right)}^3}\right)\\ & =& 0\end{array}$

From the above equations, E is conservative.

Find Electric field in rectangular coordinate.

Write the rectangular coordinate.

$$\begin{gathered} r={\left(x^2+y^2+z^2\right)}^{1/2}, \\ \mathrm{r}=x\mathrm{i}+y\mathrm{j}+z\mathrm{k} \\ \varphi =\frac{q}{{\left(x^2+y^2+z^2\right)}^{1/2}} \end{gathered}$$

Use the formula .$\mathrm{E}=-\nabla \varphi$

localid="1664276297858" $$\begin{gathered} \mathrm{E}=-\partial /\partial x\left(\frac{q}{{\left(x^2+y^2+z^2\right)}^{1/2}}\right)\mathrm{i}-\partial /\partial y\left(\frac{q}{{\left(x^2+y^2+z^2\right)}^{1/2}}\right)\mathrm{j}-\partial /\partial z\left(\frac{q}{{\left(x^2+y^2+z^2\right)}^{1/2}}\right)\mathrm{k} \\ \mathrm{E}=-\left(\frac{-qx{\left(x^2+y^2+z^2\right)}^{-1/2}}{\left(x^2+y^2+z^2\right)}\right)\mathrm{i}-\left(\frac{-qy{\left(x^2+y^2+z^2\right)}^{-1/2}}{\left(x^2+y^2+z^2\right)}\right)\mathrm{j}-\left(\frac{-qz{\left(x^2+y^2+z^2\right)}^{-1/2}}{\left(x^2+y^2+z^2\right)}\right)\mathrm{k} \\ \mathrm{E}=q(x\mathrm{i}+y\mathrm{j}+z\mathrm{k})\frac{1}{{\left(x^2+y^2+z^2\right)}^{3/2}} \\ \mathrm{E}=\frac{q\mathrm{r}}{r^3} \end{gathered}$$

Find Electric field in cylindrical coordinate

Write the cylindrical coordinate.

$$\begin{gathered} x=s\cos \varphi , \\ y=s\sin \varphi , \\ z=z \end{gathered}$$

Write another value.

$$\begin{gathered} r={\left(s^2{\cos }^2\varphi +s^2{\sin }^2\varphi +z^2\right)}^{1/2} \\ r={\left(s^2+z^2\right)}^{1/2} \\ \varphi =\frac{q}{{\left(s^2+z^2\right)}^{1/2}} \end{gathered}$$

Use the formula .$\mathrm{E}=-\nabla \varphi$

$$\begin{gathered} \mathrm{E}=-\partial /\partial s\left(\frac{q}{{\left(s^2+z^2\right)}^{1/2}}\right){\mathrm{e}}_s-\frac{1}{s}\partial /\partial \varphi \left(\frac{q}{{\left(s^2+z^2\right)}^{1/2}}\right){\mathrm{e}}_{\varphi }-\partial /\partial z\left(\frac{q}{{\left(s^2+z^2\right)}^{1/2}}\right){\mathrm{e}}_z \\ \mathrm{E}=-\left(\frac{-qs{\left(s^2+z^2\right)}^{-1/2}}{\left(s^2+z^2\right)}\right){\mathrm{e}}_s-\frac{1}{s}\left(0\right){\mathrm{e}}_{\varphi }-\partial /\partial z\left(\frac{-qz{\left(s^2+z^2\right)}^{-1/2}}{\left(s^2+z^2\right)}\right){\mathrm{e}}_z \\ \mathrm{E}=\frac{q\left(s{\mathrm{e}}_s+z{\mathrm{e}}_z\right)}{{\left(s^2+z^2\right)}^{3/2}} \end{gathered}$$

But

$$\begin{gathered} {\mathrm{e}}_s=cos\varphi \mathrm{i}+sin\varphi \mathrm{j} \\ {\mathrm{e}}_z=\mathrm{k} \end{gathered}$$

Hence .$\left(s{\mathrm{e}}_s+z{\mathrm{e}}_z\right)=scos\varphi \mathrm{i}+ssin\varphi \mathrm{j}+z\mathrm{k}$

Use the above equation.

$\begin{array}{rcl}\left(s{\mathrm{e}}_s+z{\mathrm{e}}_z\right)& =& \frac{x}{cos\varphi }cos\varphi \mathrm{i}+\frac{y}{sin\varphi }sin\varphi \mathrm{j}+z\mathrm{k}\\ & =& (x\mathrm{i}+y\mathrm{j}+z\mathrm{k})\\ & =& \mathrm{r}\end{array}$

Write further.

$\mathrm{E}=\frac{\mathrm{r}}{r^3\text{'}\text{'}\text{'}}$

Hence, the electric field is conservative.$\nabla \times \mathrm{E}=0$

The electric field is mentioned below.

$$\begin{gathered} \mathrm{E}=q(x\mathrm{i}+y\mathrm{j}+z\mathrm{k})\frac{1}{{\left(x^2+y^2+z^2\right)}^{3/2}} \\ \mathrm{E}=\frac{q\left(s{\mathrm{e}}_s+z{\mathrm{e}}_z\right)}{{\left(s^2+z^2\right)}^{3/2}} \end{gathered}$$