Chapter 23, Problem 42P

(a)

To determine

The horizontal and vertical component of electric field at a point $P$ on the $y$ axis at a distance $d$ from origin.

Answer to Problem 42P

The horizontal component of electric field at a distance $r$ due to a change $dQ$ is $-\left(\frac{kQ}{L}\right)\left(\frac{1}{{\left(d^2+L^2\right)}^{1/2}}-\frac{1}{d}\right)\widehat{i}$ and the vertical component of electric field at a distance $r$ due to a change $dQ$ is $\left(\frac{kQ}{d{\left(d^2+L^2\right)}^{1/2}}\right)\widehat{j}$.

Explanation of Solution

Write the expression for the electric field at a distance $r$ due to a charge $dQ$.

    $dE=\frac{kdQ}{r^2}$

Here, $dQ$ is a small elemental charge, $r$ is the radius and $k$ is the Coulomb’s constant.

Write the value for the Coulomb’s constant.

    $k=8.9876\times {\text{10}}^9\text{N}.{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$

The following figure represents the components of an electric field at a point $P$, due to a small elemental charge of magnitude $dQ$, considered in a small segment of rod of length $dl$.

Figure-(1)

Write the expression for uniformly distributed charge along the length of the rod.

    $\lambda =\frac{Q}{L}$

Here, $Q$ is the total charge, $L$ is the length of the rod.

Write the expression for the charge on a small elemental length of the rod.

    $dQ=\lambda dl$

Here, $\lambda$ is the charge per unit length and $dl$ is the small elemental length of the rod.

Write the sine expression for the right angled triangle.

    $\sin \theta =\frac{d}{r}$

Substitute $\sqrt{d^2+l^2}$ for $r$ in the above equation.

    $\sin \theta =\frac{d}{\sqrt{d^2+l^2}}$

Write the cosine expression for the right angled triangle.

    $\cos \theta =\frac{l}{r}$

Substitute $\sqrt{d^2+l^2}$ for $r$ in the above equation.

    $\cos \theta =\frac{l}{\sqrt{d^2+l^2}}$

Write the expression for the horizontal component of the electric field at a distance $r$ due to a charge $dQ$.

    $d{E}_{\text{x}}=\frac{kdQ}{r^2}\left(\cos \theta \right)$

Integrating the above equation between the limits $l=0$ to $l=L$.

    ${\displaystyle \int d{E}_{\text{x}}}={\displaystyle \underset{0}{\overset{L}{\int }}\frac{kdQ}{r^2}\left(\cos \theta \right)}\widehat{i}$                                                                                               (I)

Write the expression for the vertical component of electric field at a distance $r$ due to a charge $dQ$.

    $d{E}_{\text{y}}=\frac{kdQ}{r^2}\left(\sin \theta \right)$

Integrating the above equation between the limits $l=0$ to $l=L$.

    ${\displaystyle \int d{E}_{\text{y}}}={\displaystyle \underset{0}{\overset{L}{\int }}\frac{kdQ}{r^2}\left(\sin \theta \right)}\widehat{j}$                                                                                             (II)

Calculate the horizontal component of electric field at a distance $r$ due to a charge $dQ$.

Substitute $\frac{l}{r}$ for $\cos \theta$ and $\lambda dl$ for $dQ$ in equation (I) to calculate ${\displaystyle \int d{E}_{\text{x}}}$.

    $\begin{array}{c}{\displaystyle \int d{E}_{\text{x}}}={\displaystyle \underset{0}{\overset{L}{\int }}\frac{k\lambda dl}{r^2}\left(\frac{l}{r}\right)}\widehat{i}\\ =k\lambda {\displaystyle \underset{0}{\overset{L}{\int }}\frac{dl}{r^2}\left(\frac{l}{r}\right)}\widehat{i}\\ =k\lambda {\displaystyle \underset{0}{\overset{L}{\int }}\frac{l\times dl}{r^3}}\widehat{i}\end{array}$                                                                                               (III)

Write the Pythagoras theorem to calculate the value of $r$.

    $r^2=d^2+l^2$

    $r={\left(d^2+l^2\right)}^{1/2}$

Taking the cube of both sides in the above equation.

    $r^3={\left(d^2+l^2\right)}^{3/2}$

Substitute ${\left(d^2+l^2\right)}^{3/2}$ for $r^3$ in equation (III), to calculate $E_{\text{x}}$.

    $E_{\text{x}}=k\lambda {\displaystyle \underset{0}{\overset{L}{\int }}\frac{l\times dl}{{\left(d^2+l^2\right)}^{3/2}}}\widehat{i}$

As per the formula ${\displaystyle \int \frac{xdx}{{\left(x^2+a^2\right)}^{1/2}}}=\frac{-1}{{\left(x^2+a^2\right)}^{1/2}}$, thus the above equation is written as,

    $\begin{array}{c}{E}_{\text{x}}=k\lambda {\left(-\frac{1}{{\left(d^2+l^2\right)}^{1/2}}\right)}_0^L\widehat{i}\\ =-k\lambda \left(\frac{1}{{\left(d^2+L^2\right)}^{1/2}}-\frac{1}{{\left(d^2+0^2\right)}^{1/2}}\right)\widehat{i}\end{array}$

Substitute $\frac{Q}{L}$ for $\lambda$ in the above equation to calculate $E_{\text{x}}$.

    $\begin{array}{c}{E}_{\text{x}}=-k\left(\frac{Q}{L}\right)\left(\frac{1}{{\left(d^2+L^2\right)}^{1/2}}-\frac{1}{{\left(d^2+0^2\right)}^{1/2}}\right)\widehat{i}\\ =-k\left(\frac{Q}{L}\right)\left(\frac{1}{{\left(d^2+L^2\right)}^{1/2}}-\frac{1}{d}\right)\widehat{i}\\ =-\left(\frac{kQ}{L}\right)\left(\frac{1}{{\left(d^2+L^2\right)}^{1/2}}-\frac{1}{d}\right)\widehat{i}\end{array}$                                                            (IV)

Calculate the vertical component of electric field at a distance $r$ due to a change $dQ$.

Substitute $\frac{d}{r}$ for $\sin \theta$ and $\lambda dl$ for $dQ$ in equation (II) to calculate ${\displaystyle \int d{E}_{\text{y}}}$.

    $\begin{array}{c}{\displaystyle \int d{E}_{\text{y}}}={\displaystyle \underset{0}{\overset{L}{\int }}\frac{k\lambda dl}{r^2}\left(\frac{d}{r}\right)}\widehat{j}\\ =k\lambda {\displaystyle \underset{0}{\overset{L}{\int }}\frac{dl}{r^2}\left(\frac{d}{r}\right)}\widehat{j}\\ =k\lambda {\displaystyle \underset{0}{\overset{L}{\int }}\frac{d\times dl}{r^3}}\widehat{j}\end{array}$                                                                                               (V)

Substitute ${\left(d^2+l^2\right)}^{3/2}$ for $r^3$ in equation (V) to calculate $E_{\text{y}}$.

    $\begin{array}{c}{E}_{\text{y}}=k\lambda {\displaystyle \underset{0}{\overset{L}{\int }}\frac{d\times dl}{{\left(d^2+l^2\right)}^{3/2}}}\widehat{j}\\ =k\lambda d{\displaystyle \underset{0}{\overset{L}{\int }}\frac{dl}{{\left(d^2+l^2\right)}^{3/2}}}\widehat{j}\end{array}$

As per the formula ${\displaystyle \int \frac{dx}{{\left(x^2+a^2\right)}^{3/2}}}=\frac{x}{a^2{\left(x^2+a^2\right)}^{1/2}}$, thus, the above equation is written as,

    $\begin{array}{c}{E}_{\text{y}}=k\lambda d{\left(\frac{l}{d^2{\left(d^2+l^2\right)}^{1/2}}\right)}_0^L\widehat{j}\\ =k\lambda d\left(\frac{L}{d^2{\left(d^2+L^2\right)}^{1/2}}-\frac{0}{d^2{\left(d^2+0^2\right)}^{1/2}}\right)\widehat{j}\end{array}$

Substitute $\frac{Q}{L}$ for $\lambda$ in the above equation to calculate $E_{\text{y}}$.

    $\begin{array}{c}{E}_{\text{y}}=kd\left(\frac{Q}{L}\right)\left(\frac{L}{d^2{\left(d^2+L^2\right)}^{1/2}}-0\right)\widehat{j}\\ =kd\left(\frac{Q}{L}\right)\left(\frac{L}{d^2{\left(d^2+L^2\right)}^{1/2}}\right)\widehat{j}\\ =\left(\frac{kQ}{d{\left(d^2+L^2\right)}^{1/2}}\right)\widehat{j}\end{array}$                                                                       (VI)

Therefore, the horizontal and vertical component of electric field at a distance $r$ due to a change $dQ$ is $-\left(\frac{kQ}{L}\right)\left(\frac{1}{{\left(d^2+L^2\right)}^{1/2}}-\frac{1}{d}\right)\widehat{i}$ and $\left(\frac{kQ}{d{\left(d^2+L^2\right)}^{1/2}}\right)\widehat{j}$ respectively.

(b)

To determine

The approximate values of the horizontal and vertical components, when $d$ is greater than $l$.

Answer to Problem 42P

The horizontal and vertical component of electric field at a distance $r$ due to a charge $dQ$, if $d$ is greater than $l$ is $\left(0\widehat{i},\frac{kQ}{d^2}\widehat{j}\right)$ respectively.

Explanation of Solution

The length of the rod $L$ can be neglected in the horizontal and vertical components of the electric field, if $d$ is greater than $l$.

The horizontal component of the electric field in equation (IV) becomes,

    $\begin{array}{c}{E}_{\text{x}}=-\left(\frac{kQ}{L}\right)\left(\frac{1}{{\left(d^2+0^2\right)}^{1/2}}-\frac{1}{d}\right)\widehat{i}\\ =-\left(\frac{kQ}{L}\right)\left(\frac{1}{{\left(d^2\right)}^{1/2}}-\frac{1}{d}\right)\widehat{i}\\ =-\left(\frac{kQ}{L}\right)\left(\frac{1}{d}-\frac{1}{d}\right)\widehat{i}\\ =0\widehat{i}\end{array}$

The vertical component of the electric field in equation (VI) becomes,

    $\begin{array}{c}{E}_{\text{y}}=\left(\frac{kQ}{d{\left(d^2+0^2\right)}^{1/2}}\right)\widehat{j}\\ =\frac{kQ}{d\left[{\left(d^2\right)}^{1/2}\right]}\widehat{j}\\ =\frac{kQ}{d\left[d\right]}\widehat{j}\\ =\frac{kQ}{d^2}\widehat{j}\end{array}$

Therefore, the horizontal and vertical component of electric field at a distance $r$ due to a charge $dQ$, if $d$ is greater than $l$ is $\left(0\widehat{i},\frac{kQ}{d^2}\widehat{j}\right)$ respectively.