Chapter 2, Problem 2.26P

(a) Find the electric intensity on the z- axis produced by a cone surface that carries charge density $\theta =a,0<r<a,$ and $0<\varphi <2\pi$ in spherical coordinates. Differential area for a cone is given in spherical coordination as da=r space The cone has its vertex at the origin and Occupies the region đ�›‰= a, 0< r < a and 0< c 2z an spherical coordinates Differential area for a cone is given an spherical coordinates as da ,r sin a th d (b) Find the aboid charge on the cone (c) Specialize w result of part a so the case an which a = 90°. At which the cone flattens to a disk an the xy plane Compare this result to the answer to problem 2 ‘ (d) Show that vow part a result becomes a point charge field ben:>> a. (a) Show that vow pan a result becomes an inverse -:-dependent £ field when z<< a

To determine

(a)

The electric field intensity on the z-axis produced by a cone surface that carries a charge density in free space.

Answer to Problem 2.26P

The electric field intensity is $\frac{{\rho }_s}{2\pi {\epsilon }_0}{a}_{\rho }$.

Explanation of Solution

Given information:

   ${\rho }_s=\frac{{\rho }_0}{r}{\text{c/m}}^{\text{2}}$

   $\begin{array}{l}\text{Region}\\ \theta =a\\ 0<r<a\\ 0<\varphi <2\pi \end{array}$

Concept used:

   $dE=\frac{{\rho }_{s}dz\left(r-r\text{'}\right)}{4\pi {\epsilon }_0{\left|r-r\text{'}\right|}^3}\mathrm{...........}\left(1\right)$

Calculation:

Although

   $\begin{array}{l}r=\rho a_{\rho }\\ r\text{'}=z{a}_z\\ \text{Hence}\\ r-r\text{'}=\rho a_{\rho }-z{a}_z\end{array}$

Plugging value of $r-r\text{'}$ in equation (1)

   $\begin{array}{l}dE=\frac{{\rho }_{s}dz\text{'}\left(r-r\text{'}\right)}{4\pi {\epsilon }_0{\left|r-r\text{'}\right|}^3}\left(2\right)\\ dE=\frac{{\rho }_{s}dz\text{'}\left(\rho a_{\rho }-z\text{'}{a}_z\right)}{4\pi {\epsilon }_0{\left({\rho }^2+z{\text{'}}^2\right)}^{3/2}}\end{array}$

The differential field contribution to a point in the x - y plane is now summed by integrating the preceding differential field over a finite surface.

Since only the $\text{Eρ}$ component is present, so on simplification.

   $\begin{array}{l}dE\rho =\frac{{\rho }_s\rho dz\text{'}}{4\pi {\epsilon }_0{\left({\rho }^2+z{\text{'}}^2\right)}^{3/2}}\\ E\rho ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{{\rho }_s\rho dz\text{'}}{4\pi {\epsilon }_0{\left({\rho }^2+z{\text{'}}^2\right)}^{3/2}}}\\ \text{Integrating by change of variable},z=\rho \cot \theta ,\\ E\rho =\frac{{\rho }_s\rho }{4\pi {\epsilon }_0}{\left(\frac{z\text{'}}{{\rho }^2{\left({\rho }^2+z{\text{'}}^2\right)}^{1/2}}\right)}_{-\infty }^{\infty }\\ E\rho =\frac{{\rho }_s}{2\pi {\epsilon }_0}{a}_{\rho }\end{array}$

To determine

(b)

The total charge on the cone.

Answer to Problem 2.26P

The total charge on cone is ${\rho }_0\pi a^{3}C$.

Explanation of Solution

Given information:

   ${\rho }_s=\frac{{\rho }_0}{r}{\text{c/m}}^{\text{2}}$

   $\begin{array}{l}\text{Region}\\ \theta =a\\ 0<r<a\\ 0<\varphi <2\pi \end{array}$

   $da=r\sin adrd\varphi$

Concept used:

   $Q={\displaystyle \underset{v}{\int }{\rho }_s}{r}^2\sin \theta drd\theta d\varphi$

Calculation:

Hence    $\begin{array}{l}\text{Region}\\ \theta =a\\ 0<r<a\\ 0<\varphi <2\pi \end{array}$

Therefore, using the above formula

   $\begin{array}{l}Q={\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle {\int }_0^a\frac{{\rho }_0}{r}{r}^2\sin \theta drd\theta d\varphi }}}\\ Q=2\pi \frac{{\rho }_0{a}^3}{4}\left[1-\cos \left(\pi \right)\right]\\ Q={\rho }_0\pi a^3\text{C}\end{array}$

To determine

(c)

The electric field intensity on the z-axis produced by a cone flattered disc, when $a={90}^{°}$ for cone.

Answer to Problem 2.26P

   $\overrightarrow{E}\left(z\right)=\frac{1}{4\pi {\epsilon }_0}\left(\left(2\pi {\rho }_s\right)-\frac{\left(2\pi {\rho }_{s}z\right)}{\sqrt{R^2+z^2}}\right)\widehat{k}$

Explanation of Solution

Given information:

   ${\rho }_s=\frac{{\rho }_0}{r}{\text{c/m}}^{\text{2}}$

   $\begin{array}{l}\text{Region}\\ \theta =a={90}^{°}\\ 0<r<a\\ 0<\varphi <2\pi \end{array}$

Concept used:

The electric field for a surface charge is given by

   $\overrightarrow{E}\left(P\right)=\frac{1}{4\pi {\epsilon }_0}{\displaystyle \underset{surface}{\int }\frac{{\rho }_{s}dA}{r^2}\widehat{r}}$

the field is entirely in the vertical direction. The vertical component of the electric field is extracted by multiplying by θ, so

   $\overrightarrow{E}\left(P\right)=\frac{1}{4\pi {\epsilon }_0}{\displaystyle \underset{surface}{\int }\frac{{\rho }_{s}dA}{r^2}\cos \theta \widehat{k}}$

Calculation:

Although

   $\begin{array}{l}dA=2\pi r\text{'}dr\text{'}\\ r^2=r{\text{'}}^2+z^2\\ \cos \theta =\frac{z}{{\left(r^{\text{'}2}+z^2\right)}^{\frac{1}{2}}}\end{array}$

Plugging all values in the above formula

   $\begin{array}{l}\overrightarrow{E}\left(P\right)=\overrightarrow{E}\left(z\right)\\ =\frac{1}{4\pi {\epsilon }_0}{\displaystyle \underset{0}{\overset{R}{\int }}\frac{{\rho }_s\left(2\pi r^{\text{'}}d{r}^{\text{'}}\right)z}{{\left(r^{\text{'}}{}^2+z^2\right)}^{3/2}}\widehat{k}}\\ \text{integrating}\text{above}\text{integral}\\ \overrightarrow{E}\left(z\right)=\frac{1}{4\pi {\epsilon }_0}\left(2\pi {\rho }_{s}z\right)\left(\frac{1}{z}-\frac{1}{\sqrt{R^2+z^2}}\right)\widehat{k}\\ \overrightarrow{E}\left(z\right)=\frac{1}{4\pi {\epsilon }_0}\left(\left(2\pi {\rho }_s\right)-\frac{\left(2\pi {\rho }_{s}z\right)}{\sqrt{R^2+z^2}}\right)\widehat{k}\end{array}$

The electric field is similar, as calculated in the problem 2.23

To determine

(d)

The electric field intensity on the z-axis, when $z\gg a$.

Answer to Problem 2.26P

The electric field intensity for part a when $z\gg a$ is

Explanation of Solution

Given information:

   ${\rho }_s=\frac{{\rho }_0}{r}{\text{c/m}}^{\text{2}}$

   $\begin{array}{l}\text{Region}\\ \theta =a={90}^{°}\\ 0<r<a\\ 0<\varphi <2\pi \end{array}$

Calculation:

The development of the electric field due to the given formula as follows

   $\begin{array}{l}{E}_z=\frac{{\rho }_s}{2{\epsilon }_0}\left[1-\frac{z}{\sqrt{z^2+r^2}}\right]\\ E_z=\frac{{\rho }_s}{2{\epsilon }_0}\left[1-\frac{z}{z\sqrt{1+\frac{r^2}{z^2}}}\right]\\ E_z=\frac{{\rho }_s}{2{\epsilon }_0}\left[1-\frac{1}{1+\left(0.5\right)\frac{r^2}{z^2}}\right]\\ \text{applying}\text{given}\text{approximation}\\ z\gg a\\ E_z=\frac{{\rho }_s}{2{\epsilon }_0}\left[1-\left[1-\frac{\left(\frac{1}{2}\right)r^2}{z^2}\right]\right]\\ E_z=\frac{{\rho }_s{r}^2}{4{\epsilon }_0{z}^2}\\ \because r=a\\ E_z=\frac{{\rho }_s{a}^2}{4{\epsilon }_0{z}^2}\end{array}$

To determine

(e)

To proof:

Part a become an inverse z dependent E field when $z\ll a$.

Explanation of Solution

Given information:

   ${\rho }_s=\frac{{\rho }_0}{r}{\text{c/m}}^{\text{2}}$

   $\begin{array}{l}\text{Region}\\ \theta =a={90}^{°}\\ 0<r<a\\ 0<\varphi <2\pi \end{array}$

Calculation:

In general, the electric field expressed as

   $\begin{array}{l}{E}_z=\frac{{\rho }_s}{2{\epsilon }_0}\left[1-\frac{z}{\sqrt{z^2+r^2}}\right]\\ \text{applying}\text{z=a}\text{,the}\text{electric}\text{field}\text{reduces}\text{to}\\ E_z=\frac{{\rho }_s}{2{\epsilon }_0}\end{array}$

Hence as z reduces electric field is increasing

So in this case field, E is inverse z dependent.