Chapter 2, Problem 2.17P

A length d of the charge lies on the Z-axis infree space. The charge density on the line is $p_L=+C/m\left(0<z<d/2\right)$ and $p_L=-p_{0}C/m\left(-d/2<z<0\right)$

where P₀ is a possitive constant. (a) Find the electric field intensity E everywhere in the xy plane, expressing your result as a function of cylindrical radius p, (b) simplify your part a result for the case in radius pd, and express this result in terms of charge q=p₀d/2.

To determine

(a)

The electric field intensity in x-y plane, as a function of cylindrical radius $\rho$.

Answer to Problem 2.17P

The required electric field intensity is:

   $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \right)=-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(1-{\left(1+\frac{d^2}{4{\rho }^2}\right)}^{-1/2}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.

Explanation of Solution

Given Information:

The line charge density is given as,

   ${\rho }_L=\{\begin{array}{cc}+{\rho }_0\text{ C/m}& 0<z<d/2\\ -{\rho }_0\text{ C/m}& -d/2<z<0\end{array}$

Calculation:

Let $\left(x,y,0\right)$ be a point on x-y plane, and $\left(0,0,z\right)$ is a point on line charge. Then differential electric field,

   $d\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}=\frac{{\rho }_L\left(x{\text{<mover accent="true"><mo>a</mo> <mo>^</mo></mover>}}_x+y{\text{<mover accent="true"><mo>a</mo> <mo>^</mo></mover>}}_y-z{\text{<mover accent="true"><mo>a</mo> <mo>^</mo></mover>}}_z\right)}{4\pi {\epsilon }_0{\left(x^2+y^2+z^2\right)}^{3/2}}$

So, the electric field intensity:

$\text{<mover accent="true"><mo>E </mo> <mo>→</mo> </mover>}={\displaystyle \int d\text{<mover accent="true"><mo>E </mo> <mo>→</mo> </mover>}}={\displaystyle \underset{-d/2}{\overset{d/2}{\int }}\frac{{\rho }_L\left(x{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_x+y{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_y-z{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_z\right)}{4\pi {\epsilon }_0{\left(x^2+y^2+z^2\right)}^{3/2}}dz}$

$={\displaystyle \underset{-d/2}{\overset{0}{\int }}\frac{-{\rho }_0\left(x{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_x+y{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_y-z{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_z\right)}{4\pi {\epsilon }_0{\left(x^2+y^2+z^2\right)}^{3/2}}dz}+{\displaystyle \underset{0}{\overset{d/2}{\int }}\frac{{\rho }_0\left(x{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_x+y{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_y-z{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_z\right)}{4\pi {\epsilon }_0{\left(x^2+y^2+z^2\right)}^{3/2}}dz}$

$=\frac{{\rho }_0}{4\pi {\epsilon }_0}\left[{\displaystyle \underset{-d/2}{\overset{0}{\int }}\frac{z{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_z}{{\left(x^2+y^2+z^2\right)}^{3/2}}dz-{\displaystyle \underset{0}{\overset{d/2}{\int }}\frac{z{\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_z}{{\left(x^2+y^2+z^2\right)}^{3/2}}dz}}\right]$

$=\frac{{\rho }_0}{4\pi {\epsilon }_0}\left({\left[-\frac{1}{{\left(x^2+y^2+z^2\right)}^{1/2}}\right]}_{-d/2}^0-{\left[-\frac{1}{{\left(x^2+y^2+z^2\right)}^{1/2}}\right]}_0^{d/2}\right){\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_z$

$=-\frac{{\rho }_0}{4\pi {\epsilon }_0}\left(\left[\frac{1}{{\left(x^2+y^2\right)}^{1/2}}-\frac{1}{{\left(x^2+y^2+{\left(d/2\right)}^2\right)}^{1/2}}\right]-\left[\frac{1}{{\left(x^2+y^2+{\left(d/2\right)}^2\right)}^{1/2}}-\frac{1}{{\left(x^2+y^2\right)}^{1/2}}\right]\right){\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_z$

$=-\frac{2{\rho }_0}{4\pi {\epsilon }_0}\left(\frac{1}{{\left(x^2+y^2\right)}^{1/2}}-\frac{1}{{\left(x^2+y^2+{\left(d/2\right)}^2\right)}^{1/2}}\right){\text{<mover accent="true"><mo>a </mo> <mo>^</mo> </mover>}}_z$

So, in cylindrical coordinates,

   $\begin{array}{l}\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \right)=-\frac{2{\rho }_0}{4\pi {\epsilon }_0}\left(\frac{1}{\rho }-\frac{1}{{\left({\rho }^2+{\left(d/2\right)}^2\right)}^{1/2}}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\because x^2+y^2={\rho }^2\\ =-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(1-\frac{1}{{\left(1+{\left(d/2\rho \right)}^2\right)}^{1/2}}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\\ =-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(1-{\left(1+\frac{d^2}{4{\rho }^2}\right)}^{-1/2}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}\end{array}$

Conclusion:

The required electric field intensity is:

   $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \right)=-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(1-{\left(1+\frac{d^2}{4{\rho }^2}\right)}^{-1/2}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.

To determine

(b)

The electric field intensity in x-y plane for $\rho \gg d$.

Answer to Problem 2.17P

The required electric field intensity is,

   $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \gg d\right)=-\frac{qd}{4\pi {\epsilon }_0{\rho }^3}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.

Explanation of Solution

Given Information:

The line charge density is given as,

   ${\rho }_L=\{\begin{array}{cc}+{\rho }_0\text{ C/m}& 0<z<d/2\\ -{\rho }_0\text{ C/m}& -d/2<z<0\end{array}$

   $\begin{array}{l}\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \right)=-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(1-{\left(1+\frac{d^2}{4{\rho }^2}\right)}^{-1/2}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}\\ q={\rho }_{0}d/2\end{array}$

Calculation:

The electric field intensity,

   $\begin{array}{l}\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \gg d\right)=-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(1-{\left(1+\frac{d^2}{4{\rho }^2}\right)}^{-1/2}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\\ =-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(1-{\left(1+\frac{d^2}{4{\rho }^2}\right)}^{-1}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\because {\left(1+\frac{d^2}{4{\rho }^2}\right)}^{1/2}\approx \left(1+\frac{d^2}{4{\rho }^2}\right)\\ =-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(\frac{d^2}{4{\rho }^2+d^2}\right)\\ =-\frac{{\rho }_0}{2\pi {\epsilon }_0\rho }\left(\frac{d^2}{4{\rho }^2}\right)\because 4{\rho }^2+d^2\approx 4{\rho }^2\\ =-\frac{2q}{2\pi {\epsilon }_0\rho d}\left(\frac{d^2}{4{\rho }^2}\right)\because q={\rho }_{0}d/2\\ =-\frac{qd}{4\pi {\epsilon }_0{\rho }^3}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}\end{array}$

Conclusion:

The required electric field intensity is,

   $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \gg d\right)=-\frac{qd}{4\pi {\epsilon }_0{\rho }^3}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.