Chapter 25, Problem 71CP

(a)

To determine

The proof for $V=\frac{p{k}_{\text{e}}\cos \theta }{r^2}$ at the point $P$.

Answer to Problem 71CP

The electric potential at the point $P$ is $V=\frac{p{k}_{\text{e}}\cos \theta }{r^2}$.

Explanation of Solution

The following figure shows the diagram of the dipoles and the point $P$.

Figure-(1)

The dipole moment’s magnitude is given as.

  $p=2qa$

Here, $p$ is the magnitude of the dipole moment, $q$ is the charge and $2a$ is the separating distance between the charges, $r_1$ is the distance between the positive charge and the point $P$ and $r_2$ is the distance between the point $P$ and the negative charge.

Write the equation for the electric potential at the point $P$ due to the dipole moment.

  $\begin{array}{c}V=k_{\text{e}}\left[\frac{q}{r_1}-\frac{q}{r_2}\right]\\ =k_{\text{e}}q\left[\frac{1}{r_1}-\frac{1}{r_2}\right]\end{array}$                                                                                                       (I)

Here, $V$ is the potential, $k_{\text{e}}$ is the Coulomb’s constant,

Write the equation for the distance between the positive charge and the point $P$.

  $r^2{}_1=r^2\left(1-\frac{2a}{r}\cos \theta \right)$                                                                                              (II)

Write the equation for the distance between the negative charge and the point $P$.

  $r^2{}_2=r^2\left(1+\frac{2a}{r}\cos \theta \right)$                                                                                           (III)

Rearrange the equation (II) and (III) to calculate $\frac{1}{r_1}$ and $\frac{1}{r_2}$.

  $\frac{1}{r_1}=\frac{1}{r}{\left(1-\frac{2a}{r}\cos \theta \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ and $\frac{1}{r_2}=\frac{1}{r}{\left(1+\frac{2a}{r}\cos \theta \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Expand the above equation binomially and neglect the higher terms to calculate $\frac{1}{r_1}$ and $\frac{1}{r_2}$.

  $\frac{1}{r_1}=\frac{1}{r}\left(1-\left(\frac{-1}{2}\right)\frac{2a}{r}\cos \theta \right)$ and $\frac{1}{r_2}=\frac{1}{r}\left(1+\left(\frac{-1}{2}\right)\frac{2a}{r}\cos \theta \right)$

  $\frac{1}{r_1}=\frac{1}{r}\left(1+\frac{a}{r}\cos \theta \right)$ and $\frac{1}{r_2}=\frac{1}{r}\left(1-\frac{a}{r}\cos \theta \right)$

Substitute $\frac{1}{r}\left(1+\frac{a}{r}\cos \theta \right)$ for $\frac{1}{r_1}$ and $\frac{1}{r}\left(1-\frac{a}{r}\cos \theta \right)$ for $\frac{1}{r_2}$ in the equation (I) to calculate the electric potential.

  $\begin{array}{c}V=k_{\text{e}}q\left[\frac{1}{r}\left(1+\frac{a}{r}\cos \theta \right)-\frac{1}{r}\left(1-\frac{a}{r}\cos \theta \right)\right]\\ =k_{\text{e}}\frac{q}{r}\left[\frac{2a}{r}\cos \theta \right]\\ =k_{\text{e}}\frac{2qa}{r^2}\cos \theta \end{array}$

Substitute $p$ for $2aq$ in the above equation to calculate the electric potential.

  $V=\frac{p{k}_{\text{e}}\cos \theta }{r^2}$

Therefore, the electric potential at the point $P$ is $V=\frac{p{k}_{\text{e}}\cos \theta }{r^2}$.

(b)

To determine

The electric field’s radial component $E_{\text{r}}$ and the perpendicular component $E_{\text{θ}}$.

Answer to Problem 71CP

The electric field’s radial component is $E_{\text{r}}=\frac{2p{k}_{\text{e}}\cos \theta }{r^3}$ and the perpendicular component is $E_{\text{θ}}=\frac{p{k}_{\text{e}}\sin \theta }{r^3}$.

Explanation of Solution

Write the equation for the radial component of the electric field.

  $E_{\text{r}}=-\frac{dV}{dr}$                                                                                                               (IV)

Here, $E_{\text{r}}$ is the radial component of the electric field.

Write the equation for the perpendicular component of the electric field.

  $E_{\text{θ}}=-\left(\frac{1}{r}\right)\frac{dV}{d\theta }$                                                                                                        (V)

Here, $E_{\text{θ}}$ is the perpendicular component of the electric field.

Conclusion:

Substitute $\frac{p{k}_{\text{e}}\cos \theta }{r^2}$ for $V$ in the equation (IV) to calculate the radial component of the electric field.

  $\begin{array}{c}{E}_r=-\frac{d\left(\frac{p{k}_{\text{e}}\cos \theta }{r^2}\right)}{dr}\\ =\frac{2p{k}_{\text{e}}\cos \theta }{r^3}\end{array}$

Substitute $\frac{p{k}_{\text{e}}\cos \theta }{r^2}$ for $V$ in the equation (V) to calculate the perpendicular component of the electric field.

  $\begin{array}{c}{E}_{\text{θ}}=-\left(\frac{1}{r}\right)\frac{d\left(\frac{p{k}_{\text{e}}\cos \theta }{r^2}\right)}{d\theta }\\ =-\left(\frac{1}{r}\right)\frac{p{k}_{\text{e}}\left(-\sin \theta \right)}{r^3}\\ =\frac{p{k}_{\text{e}}\sin \theta }{r^3}\end{array}$

Therefore, the electric field’s radial component is $E_{\text{r}}=\frac{2p{k}_{\text{e}}\cos \theta }{r^3}$ and the perpendicular component is $E_{\text{θ}}=\frac{p{k}_{\text{e}}\sin \theta }{r^3}$.

(c)

To determine

Whether the results for $E_{\text{r}}$ and $E_{\text{θ}}$ hold good when $\theta =0°$ and $90°$.

Answer to Problem 71CP

Yes, the results for $E_{\text{r}}$ and $E_{\text{θ}}$ hold good when $\theta =0°$ and $\theta =90°$.

Explanation of Solution

Write the equation for $E_{\text{r}}$.

  $E_{\text{r}}=\frac{2p{k}_{\text{e}}\cos \theta }{r^3}$                                                                                                     (VI)

Write the equation for $E_{\text{θ}}$.

  $E_{\text{θ}}=\frac{p{k}_{\text{e}}\sin \theta }{r^3}$                                                                                                       (VII)

Conclusion:

Substitute $0°$ for $\theta$ in the equation (VI) to calculate $E_{\text{r}}$.

  $\begin{array}{c}{E}_{\text{r}}=\frac{2p{k}_{\text{e}}\cos 0°}{r^3}\\ =\frac{2p{k}_{\text{e}}}{r^3}\end{array}$

Substitute $0°$ for $\theta$ in the equation (VII) to calculate $E_{\text{θ}}$.

  $\begin{array}{c}{E}_{\text{θ}}=\frac{p{k}_{\text{e}}\sin 0°}{r^3}\\ =0\end{array}$

Substitute $90°$ for $\theta$ in the equation (VI) to calculate $E_{\text{r}}$.

  $\begin{array}{c}{E}_{\text{r}}=\frac{2p{k}_{\text{e}}\cos 90°}{r^3}\\ =0\end{array}$

Substitute $90°$ for $\theta$ in the equation (VII) to calculate $E_{\text{θ}}$.

    $\begin{array}{c}{E}_{\text{θ}}=\frac{p{k}_{\text{e}}\sin 90°}{r^3}\\ =\frac{p{k}_{\text{e}}}{r^3}\end{array}$

Therefore, the results for $E_{\text{r}}$ and $E_{\text{θ}}$ hold good when $\theta =0°$ and $\theta =90°$.

(d)

To determine

Whether the results for $E_{\text{r}}$ and $E_{\text{θ}}$ hold good when $r=0$.

Answer to Problem 71CP

No, the results for $E_{\text{r}}$ and $E_{\text{θ}}$ do not hold good when $r=0$.

Explanation of Solution

Write the equation for $E_{\text{r}}$.

  $E_{\text{r}}=\frac{2p{k}_{\text{e}}\cos \theta }{r^3}$                                                                                                    (VIII)

Write the equation for $E_{\text{θ}}$.

  $E_{\text{θ}}=\frac{p{k}_{\text{e}}\sin \theta }{r^3}$                                                                                                         (IX)

Conclusion:

Substitute $0$ for $r$ in the equation (VIII) and (IX) to calculate $E_{\text{r}}$.

  $\begin{array}{c}{E}_{\text{r}}=\frac{2p{k}_{\text{e}}\cos \theta }{0^3}\\ =\infty \end{array}$

Substitute $0$ for $r$ in the equation (IX) to calculate $E_{\text{θ}}$.

    $\begin{array}{c}{E}_{\text{θ}}=\frac{p{k}_{\text{e}}\sin \theta }{0^3}\\ =\infty \end{array}$

Therefore, the results for $E_{\text{r}}$ and $E_{\text{θ}}$ do not hold good when $r=0$.

(e)

To determine

The expression for the electric potential in terms of the Cartesian coordinates.

Answer to Problem 71CP

The expression for the electric potential in terms of the Cartesian coordinates is $V=\frac{p{k}_{\text{e}}y}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$.

Explanation of Solution

Write the equation for the electric potential.

  $V=\frac{p{k}_{\text{e}}\cos \theta }{r^2}$

Conclusion:

Substitute $x^2+y^2$ for $r^2$ and $\frac{y}{\sqrt{x^2+y^2}}$ for $\cos \theta$ in the equation above to calculate the electric potential.

  $\begin{array}{c}V=\frac{p{k}_{\text{e}}\left(\frac{y}{\sqrt{x^2+y^2}}\right)}{{\left(x^2+y^2\right)}^2}\\ =\frac{p{k}_{\text{e}}y}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$

Therefore, the expression for the electric potential in terms of the Cartesian coordinates is $V=\frac{p{k}_{\text{e}}y}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$.

(f)

To determine

The $x$ component and the $y$ component of the electric field.

Answer to Problem 71CP

The $x$ component and the $y$ component of the electric field are $E_{\text{x}}=3p{k}_{\text{e}}\frac{xy}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ and $E_{\text{y}}=p{k}_{\text{e}}\times \frac{\left(2y^2-x^2\right)}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$.

Explanation of Solution

Write the equation for the $x$ component of the electric field.

  $E_x=-\frac{dV}{dx}$                                                                                                                (X)

Here, $E_{\text{x}}$ is the $x$ component of the electric field.

Write the equation for the $y$ component of the electric field.

  $E_x=-\frac{dV}{dy}$                                                                                                              (XI)

Here, $E_{\text{y}}$ is the $y$ component of the electric field.

Conclusion:

Substitute $\frac{p{k}_{\text{e}}y}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ for $V$ in the equation (X) to calculate the $x$ component of the electric field.

  $\begin{array}{c}{E}_x=-\frac{d\left[\frac{p{k}_{\text{e}}y}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}{dx}\\ =3p{k}_{\text{e}}\frac{xy}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$

Substitute $\frac{p{k}_{\text{e}}y}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ for $V$ in the equation (XI) to calculate the $y$ component of the electric field.

  $\begin{array}{c}{E}_{\text{y}}=-\frac{d\left[\frac{p{k}_{\text{e}}y}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}{dy}\\ =p{k}_{\text{e}}\frac{\left(2y^2-x^2\right)}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}$

Therefore, the $x$ component and the $y$ component of the electric field are $E_{\text{x}}=3p{k}_{\text{e}}\frac{xy}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ and $E_{\text{y}}=p{k}_{\text{e}}\times \frac{\left(2y^2-x^2\right)}{{\left(x^2+y^2\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$.