Chapter 25, Problem 23P

(a)

To determine

The finite value(s) of $x$ for which the electric field is zero.

Answer to Problem 23P

$-4.83\text{m}$ is the point where the net electric field will be zero.

Explanation of Solution

The following figure shows the direction of the electric field at point $P$ due to charges at points $O$ and $A$; where, the point $O$ is the origin.

Figure-(1)

Consider the distance between point $O$ and $P$ as $r$ and distance between point $O$ and $A$ as $x$ as shown in the Figure (1).

Write the expression for electric field.

    $\overrightarrow{E}=k_{\text{e}}\times \frac{q}{d^2}\left(\widehat{i}\right)$ (I)

Here, $\overrightarrow{E}$ is the electric field, $k_{\text{e}}$ is the Coulomb’s constant, $q$ is the charge, $d$ is the distance from points and $\widehat{i}$ is the field direction.

Substitute, $+q$ for $q$ and $r$ for $d$ in Equation (I) to calculate $\overrightarrow{E_1}$.

    $\overrightarrow{E_1}=k_{\text{e}}\times \frac{q}{r^2}\left(\widehat{i}\right)$

Substitute $-2q$ for $q$ and $r-x$ for $d$ Equation (I) to calculate $\overrightarrow{E_2}$.

    $\overrightarrow{E_2}=k_{\text{e}}\times \frac{-2q}{{\left(r-x\right)}^2}\left(-\widehat{i}\right)$

The net electric field at point $P$ due to the charges $+q$ and $-2q$ is zero, therefore.

    $\begin{array}{c}\overrightarrow{E_{\text{net}}}=\overrightarrow{E_1}+\overrightarrow{E_2}\\ =0\end{array}$                                                                                                       (II)

Substitute $k_{\text{e}}\times \frac{q}{d^2}\left(\widehat{i}\right)$ for $\overrightarrow{E_1}$ and $k_{\text{e}}\times \frac{-2q}{{\left(r-x\right)}^2}\left(-\widehat{i}\right)$ for $\overrightarrow{E_2}$ in Equation (II) to calculate $\overrightarrow{E_{\text{net}}}$.

    $\begin{array}{c}{k}_{\text{e}}\times \frac{-2q}{{\left(r-x\right)}^2}\left(-\widehat{i}\right)+k_{\text{e}}\times \frac{q}{r^2}\left(\widehat{i}\right)=0\\ \frac{-2q}{{\left(r-x\right)}^2}\left(-\widehat{i}\right)+\frac{q}{r^2}\left(\widehat{i}\right)=0\\ \frac{2q}{{\left(r-x\right)}^2}\left(\widehat{i}\right)=\frac{q}{r^2}\left(\widehat{i}\right)\\ \frac{2q}{{\left(r-x\right)}^2}=\frac{q}{r^2}\end{array}$

Further solve the above equation.

    $\begin{array}{c}2\times r^2={\left(r-x\right)}^2\\ \sqrt{2\times r^2}=\left(r-x\right)\\ \sqrt{2}\times r=\left(r-x\right)\\ \left(\sqrt{2}\times r\right)-r=-x\end{array}$

Further solve the above equation.

    $\begin{array}{c}r\left(\sqrt{2}-1\right)=-x\\ r=\frac{-x}{\sqrt{2}-1}\end{array}$                                                                                               (III)

Conclusion:

Substitute $2.00\text{m}$ for $x$ in Equation (III) to calculate $r$.

    $\begin{array}{c}r=\frac{-2.00\text{m}}{\sqrt{2}-1}\\ =-4.83\text{m}\end{array}$

Therefore, $-4.83\text{m}$ is the point where, the net electric field will be zero.

(b)

To determine

The finite value(s) of $x$ for which the electric potential is zero.

Answer to Problem 23P

$-2\text{m}$ and $0.667\text{m}$ is the point where the net electric potential will be zero.

Explanation of Solution

The following figure shows the net electric field at point $P$ due to the charges at points $O$ and $A$; where, the point $O$ is the origin.

Figure-(2)

Consider the distance between point $O$ and $P$ as $r$ and the distance between points $O$ and $A$ as $x$ as shown in figure (2).

Write the expression for the electric field.

    $V=k_{\text{e}}\frac{q}{d}$ (IV)

Here, $V$ is the electric potential, $k_{\text{e}}$ is the Coulomb’s constant, $q$ is the charge, $d$ is the distance from points and $\widehat{i}$ is the field direction.

Substitute, $+q$ for $q$ and $r$ for $d$ in Equation (I) to calculate $V_1$.

    $V_1=k_{\text{e}}\times \frac{q}{r}$

Substitute $-2q$ for $q$ and $r-x$ for $d$ Equation (I) to calculate $V_2$.

    $V_2=k_{\text{e}}\times \frac{-2q}{\left(r-x\right)}$

The net electric potential at point $P$ due to the charges $+q$ and $-2q$ is zero, therefore.

    $\begin{array}{c}{V}_{\text{net}}=V_1+V_2\\ =0\end{array}$                                                                                                            (V)

Substitute $k_{\text{e}}\times \frac{q}{r}$ for $V_1$ and $k_{\text{e}}\times \frac{-2q}{\left(r-x\right)}$ for $V_2$ in Equation (V) to calculate $r$.

    $\begin{array}{c}{k}_{\text{e}}\times \frac{-2q}{\left(r-x\right)}+k_{\text{e}}\times \frac{q}{r}=0\\ \frac{-2q}{\left(r-x\right)}+\frac{q}{r}=0\\ \frac{2q}{\left(r-x\right)}=\frac{q}{r}\\ \frac{2}{\left(r-x\right)}=\frac{1}{r}\end{array}$

Further solve the above equation.

    $\begin{array}{c}2r=r-x\\ r=-x\end{array}$                                                                                                             (VI)

Substitute $-2q$ for $q$ and $x-r$ for $d$ Equation (I) to calculate $V_2$.

    $V_2=k_{\text{e}}\times \frac{-2q}{\left(x-r\right)}$

Substitute $k_{\text{e}}\times \frac{q}{r}$ for $V_1$ and $k_{\text{e}}\times \frac{-2q}{\left(x-r\right)}$ for $V_2$ in Equation (II) to calculate $r$.

    $\begin{array}{c}{k}_{\text{e}}\times \frac{-2q}{\left(x-r\right)}+k_{\text{e}}\times \frac{q}{r}=0\\ \frac{-2q}{\left(x-r\right)}+\frac{q}{r}=0\\ \frac{2q}{\left(x-r\right)}=\frac{q}{r}\\ \frac{2}{\left(x-r\right)}=\frac{1}{r}\end{array}$

Further solve the above equation.

    $\begin{array}{c}2r=x-r\\ 3r=x\\ r=\frac{x}{3}\end{array}$                                                                                                               (VII)

Conclusion:

Substitute $2.00\text{m}$ for $x$ in Equation (VI) to calculate $r_1$.

    $r_1=-2.00\text{m}$

Substitute $2.00\text{m}$ for $x$ in Equation (VI) to calculate $r_2$.

    $\begin{array}{c}{r}_2=\frac{2.00\text{m}}{3}\\ =0.667\text{m}\end{array}$

Therefore, $-2\text{m}$ and $0.667\text{m}$ is the point where the net electric potential will be zero.