Chapter 2, Problem 2.7P

Point charges of equal magnitude but of opposite sign are positioned the charge +q at z = -d/2 and charge -q at z= -d/2 The charges in this configuration form an e1ectric dispole (a) Find the electric field intensity E everywhere on the z-axis is (b) Evaluate pan a result at the origin
(c) Find the electric geld intensity everywhere on the zy plane, expressing your result as a function of radius p in cylindrical coordinates, (d) Evaluate your pan c result at the origin (e) Simplify your part c result for the case in which p >> d.

To determine

(a)

The electric field intensity on z -axis.

Answer to Problem 2.7P

The required electric field intensity is:

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

Explanation of Solution

Given Information:

The point charge $+q$ is at $z=+d/2$ and $-q$ is at $z=-d/2$.

Calculation:

Let $\left(0,0,z\right)$ be a point on z axis. Then electric field due to two point charges,

   $\begin{array}{l}\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(z\right)=\frac{q\left(z{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z-\left(d/2\right){\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z\right)}{4\pi {\epsilon }_0{\left(z-\left(d/2\right)\right)}^3}-\frac{q\left(z{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z+\left(d/2\right){\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z\right)}{4\pi {\epsilon }_0{\left(z+\left(d/2\right)\right)}^3}\\ =\frac{q}{4\pi {\epsilon }_0}\left[\frac{z{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z-\left(d/2\right){\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z}{{\left|\left(z-\left(d/2\right)\right)\right|}^3}-\frac{z{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z+\left(d/2\right){\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z}{{\left|\left(z+\left(d/2\right)\right)\right|}^3}\right]\\ =\frac{q{\text{<mover accent="true"><mo>a <mo>^</mo></mo></mover>}}_z}{4\pi {\epsilon }_0}\left[\frac{1}{{\left(z-\left(d/2\right)\right)}^2}-\frac{1}{{\left(z+\left(d/2\right)\right)}^2}\right]\\ =\frac{q{\text{<mover accent="true"><mo>a <mo>^</mo></mo></mover>}}_z}{4\pi {\epsilon }_0{z}^2}\left[{\left(1-\frac{d}{2z}\right)}^{-2}-{\left(1+\frac{d}{2z}\right)}^{-2}\right]\text{ V/m}\end{array}$

Conclusion:

The required electric field intensity is:

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

To determine

(b)

The electric field intensity at origin.

Answer to Problem 2.7P

The required electric field intensity is,

   $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(z=0\right)=-\frac{2q}{\pi {\epsilon }_0{d}^2}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{V/m}$.

Explanation of Solution

Given Information:

The point charge $+q$ is at $z=+d/2$ and $-q$ is at $z=-d/2$.

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

Calculation:

The electric field at origin,

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

Conclusion:

The required electric field intensity is $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(z=0\right)=-\frac{2q}{\pi {\epsilon }_0{d}^2}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.

To determine

(c)

The electric field intensity on x-y plane as a function of radius in cylindrical coordinates.

Answer to Problem 2.7P

The required electric field intensity is $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \right)=-\frac{qd}{4\pi {\epsilon }_0{\left({\rho }^2+{\left(d/2\right)}^2\right)}^{3/2}}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.

Explanation of Solution

Given Information:

The point charge $+q$ is at $z=+d/2$ and $-q$ is at $z=-d/2$.

Calculation:

Let $\left(x,y,0\right)$ be a point on x-y plane. Then electric field due to two-point charges,

   $\begin{array}{l}\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(x,y\right)=\frac{q\left(x{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_x+z{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_y-\left(d/2\right){\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z\right)}{4\pi {\epsilon }_0{\left(x^2+y^2+{\left(d/2\right)}^2\right)}^{3/2}}-\frac{q\left(x{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_x+z{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_y+\left(d/2\right){\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z\right)}{4\pi {\epsilon }_0{\left(x^2+y^2+{\left(d/2\right)}^2\right)}^{3/2}}\\ =\frac{q}{4\pi {\epsilon }_0}\left[\frac{x{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_x+z{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_y-\left(d/2\right){\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z}{{\left(x^2+y^2+{\left(d/2\right)}^2\right)}^{3/2}}-\frac{x{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_x+z{\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_y-\left(d/2\right){\text{<mover accent="true"><mo> a <mo>^</mo> </mo></mover>}}_z}{{\left(x^2+y^2+{\left(d/2\right)}^2\right)}^{3/2}}\right]\\ =\frac{q}{4\pi {\epsilon }_0}\left(\frac{-d}{{\left(x^2+y^2+{\left(d/2\right)}^2\right)}^{3/2}}\right){\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\end{array}$

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

Conclusion:

The required electric field intensity is $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \right)=-\frac{qd}{4\pi {\epsilon }_0{\left({\rho }^2+{\left(d/2\right)}^2\right)}^{3/2}}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.

To determine

(d)

The electric field intensity at origin.

Answer to Problem 2.7P

The required electric field intensity is $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho =0\right)=-\frac{2q}{\pi {\epsilon }_0{d}^2}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.

Explanation of Solution

Given Information:

The point charge $+q$ is at $z=+d/2$ and $-q$ is at $z=-d/2$.

   $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \right)=-\frac{qd}{4\pi {\epsilon }_0{\left({\rho }^2+{\left(d/2\right)}^2\right)}^{3/2}}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$

Calculation:

The electric field intensity at origin,

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

Conclusion:

The required electric field intensity is $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho =0\right)=-\frac{2q}{\pi {\epsilon }_0{d}^2}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$.

To determine

(e)

The electric field intensity on x-y plane when $\rho \gg d$.

Answer to Problem 2.7P

The required electric field intensity is,

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

Explanation of Solution

Given Information:

The point charge $+q$ is at $z=+d/2$ and $-q$ is at $z=-d/2$.

   $\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}\left(\rho \right)=-\frac{qd}{4\pi {\epsilon }_0{\left({\rho }^2+{\left(d/2\right)}^2\right)}^{3/2}}{\text{<mover accent="true"><mo>a</mo><mo>^</mo></mover>}}_z\text{ V/m}$

Calculation:

When $\rho \gg d$ , the electric field intensity,

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

Conclusion:

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