Chapter 4, Problem 3P

(a)

To determine

The electric field $E$ at $\left(0,0,0\right)$.

Explanation of Solution

Given:

The first charge $\left(Q_1\right)$ is $Q$.

The second charge $\left(Q_2\right)$ is $-Q$.

Location of the $Q_1$ charge $\left(r_1\right)$ is $\left(a,0,0\right)$.

Location of the $Q_2$ charge $\left(r_2\right)$ is $\left(-a,0,0\right)$.

Location of the $E$ electric field $\left(r\right)$ is $\left(0,0,0\right)$.

Calculation:

Calculate the location of the electric field $\left(E\right)$ using the relation.

$\begin{array}{c}r=xi+yj+zk\\ =\left(0\right)i+\left(0\right)j+\left(0\right)k\\ =0\end{array}$

Calculate the location of the $Q_1$ charge in vector form using the relation.

$\begin{array}{c}{r}_1=x_{1}i+y_{1}j+z_{1}k\\ =\left(\text{a}\right)i+\left(0\right)j+\left(0\right)k\\ =\text{a}i\end{array}$

Calculate the location of the $Q_2$ charge in vector form using the relation.

$\begin{array}{c}{r}_2=x_{2}i+y_{2}j+z_{2}k\\ =\left(-\text{a}\right)i+\left(0\right)j+\left(0\right)k\\ =-\text{a}i\end{array}$

Calculate the electric field $\left(E\right)$.using the relation.

  $E=\frac{Q_1\left(r-r_1\right)}{4\pi {\epsilon }_0{\left|r-r_1\right|}^3}+\frac{Q_2\left(r-r_2\right)}{4\pi {\epsilon }_0{\left|r-r_2\right|}^3}$

  $\begin{array}{c}E=\frac{Q\left[0-\left(\text{a}i\right)\right]}{4\pi {\epsilon }_0{\left|0-\left(\text{a}i\right)\right|}^3}+\frac{\left(-Q\right)\left[0-\left(-\text{a}i\right)\right]}{4\pi {\epsilon }_0{\left|0-\left(-\text{a}i\right)\right|}^3}\\ =-\frac{Q\text{a}i}{4\pi {\epsilon }_0\left({\text{a}}^3\right)}-\frac{Q\text{a}i}{4\pi {\epsilon }_0\left({\text{a}}^3\right)}\\ =\frac{-2Q\text{a}i}{4\pi {\epsilon }_0{\text{a}}^3}\\ =\frac{-Q}{2\pi {\epsilon }_0{\text{a}}^2}{a}_x\end{array}$

Thus, the electric field $E$ at $\left(0,0,0\right)$ is $\frac{-Q}{2\pi {\epsilon }_0{\text{a}}^2}{a}_x$.

(b)

To determine

The electric field $E$ at $\left(0,a,0\right)$.

Explanation of Solution

Given:

Location of the $E$ electric field $\left(r\right)$ is $\left(0,a,0\right)$.

Calculation:

Calculate the location of the $E$ electric field using the relation.

$\begin{array}{c}r=xi+yj+zk\\ =\left(0\right)i+\left(\text{a}\right)j+\left(0\right)k\\ =\text{a}j\end{array}$

Calculate the electric field $\left(E\right)$.using the relation.

  $E=\frac{Q_1\left(r-r_1\right)}{4\pi {\epsilon }_0{\left|r-r_1\right|}^3}+\frac{Q_2\left(r-r_2\right)}{4\pi {\epsilon }_0{\left|r-r_2\right|}^3}$

  $\begin{array}{c}E=\frac{Q\left[\text{a}j-\left(\text{a}i\right)\right]}{4\pi {\epsilon }_0{\left|\text{a}j-\left(\text{a}i\right)\right|}^3}+\frac{\left(-Q\right)\left[\text{a}j-\left(-\text{a}i\right)\right]}{4\pi {\epsilon }_0{\left|\text{a}j-\left(-\text{a}i\right)\right|}^3}\\ =\frac{Q\text{a}\left(j-i\right)}{4\pi {\epsilon }_0\left[{\text{a}}^3{\left(\sqrt{1^2+{\left(-1\right)}^2}\right)}^3\right]}-\frac{Q\text{a}\left(j+i\right)}{4\pi {\epsilon }_0\left[{\text{a}}^3{\left(\sqrt{1^2+{\left(-1\right)}^2}\right)}^3\right]}\\ =\frac{Q\text{a}\left(j-i\right)}{4\pi {\epsilon }_0{\text{a}}^{3}2\sqrt{2}}-\frac{Q\text{a}\left(j+i\right)}{4\pi {\epsilon }_0{\text{a}}^{3}2\sqrt{2}}\\ =\frac{Q\text{a}j}{4\pi {\epsilon }_0{\text{a}}^{3}2\sqrt{2}}-\frac{Q\text{a}i}{4\pi {\epsilon }_0{\text{a}}^{3}2\sqrt{2}}-\frac{Q\text{a}j}{4\pi {\epsilon }_0{\text{a}}^{3}2\sqrt{2}}-\frac{\text{Qa}i}{4\pi {\epsilon }_0{\text{a}}^{3}2\sqrt{2}}\end{array}$

  $\begin{array}{c}E=\frac{-\text{2}Q\text{a}i}{4\pi {\epsilon }_0{\text{a}}^{3}2\sqrt{2}}\\ =\frac{-Q}{4\sqrt{2}\pi {\epsilon }_0{\text{a}}^2}{\text{a}}_x\end{array}$

Thus, the electric field $E$ at $\left(0,a,0\right)$ is $\frac{-Q}{4\sqrt{2}\pi {\epsilon }_0{\text{a}}^2}{\text{a}}_x$.

(c)

To determine

The electric field $E$ at $\left(a,0,a\right)$.

Explanation of Solution

Given:

Location of the $E$ electric field $\left(r\right)$ is $\left(a,0,a\right)$.

Calculation:

Calculate the location of the $E$ electric field using the relation.

$\begin{array}{c}r=xi+yj+zk\\ =\left(\text{a}\right)i+\left(0\right)j+\left(\text{a}\right)k\\ =\text{a}i+\text{a}k\end{array}$

Calculate the electric field $\left(E\right)$.using the relation.

    $E=\frac{Q_1\left(r-r_1\right)}{4\pi {\epsilon }_0{\left|r-r_1\right|}^3}+\frac{Q_2\left(r-r_2\right)}{4\pi {\epsilon }_0{\left|r-r_2\right|}^3}$

  $\begin{array}{c}E=\frac{Q\left[\left(\text{a}i+\text{a}k\right)-\left(\text{a}i\right)\right]}{4\pi {\epsilon }_0{\left|\left(\text{a}i+\text{a}k\right)-\left(\text{a}i\right)\right|}^3}+\frac{\left(-Q\right)\left[\left(\text{a}i+\text{a}k\right)-\left(-\text{a}i\right)\right]}{4\pi {\epsilon }_0{\left|\left(\text{a}i+\text{a}k\right)-\left(-\text{a}i\right)\right|}^3}\\ =\frac{Q\text{a}\left(k\right)}{4\pi {\epsilon }_0{\left(\text{a}k\right)}^3}-\frac{Q\text{a}\left(2i+k\right)}{4\pi {\epsilon }_0{\left[\text{a}\left(2i+k\right)\right]}^3}\\ =\frac{Q\text{a}k}{4\pi {\epsilon }_0{\text{a}}^3}-\frac{Q\text{a}k}{4\pi {\epsilon }_0{\text{a}}^3{\left(\sqrt{2^2+1^2}\right)}^3}-\frac{\text{2}Q\text{a}i}{4\pi {\epsilon }_0{\text{a}}^3{\left(\sqrt{2^2+1^2}\right)}^3}\\ =\frac{Q\text{a}k}{4\pi {\epsilon }_0{\text{a}}^3}-\frac{Q\text{a}k}{20\pi {\epsilon }_0{\text{a}}^3\sqrt{5}}-\frac{\text{2}Q\text{a}i}{20\pi {\epsilon }_0{\text{a}}^3\sqrt{5}}\end{array}$

  $\begin{array}{c}E=\frac{Q\text{a}k}{4\pi {\epsilon }_0{\text{a}}^3}\left(1-\frac{1}{5\sqrt{5}}\right)-\frac{Q\text{a}i}{10\pi {\epsilon }_0{\text{a}}^3\sqrt{5}}\\ =\frac{Q}{10\sqrt{5}\pi {\epsilon }_0{\text{a}}^2}\left[-i+5.1k\right]\\ =\frac{Q}{10\sqrt{5}\pi {\epsilon }_0{\text{a}}^2}\left[-a_x+5.1a_z\right]\end{array}$

Thus, the electric field $E$ at $\left(a,0,a\right)$ is $\frac{Q}{10\sqrt{5}\pi {\epsilon }_0{\text{a}}^2}\left[-a_x+5.1a_z\right]$.