Chapter 17, Problem 19P

To determine

The net electric field and electric potential at the center of the square.

Answer to Problem 19P

The net electric field is $5.4\times {10}^8{\text{NC}}^{\text{-1}}$ towards the line joining the center and the corner $c$ and the electric potential is $0$.

Explanation of Solution

The charges at the corners of the square are $+9.0\text{ μC}$, $-3.0\text{μC}$, $-3.0\text{μC}$ and $-3.0\text{μC}$ at the corners $a$, $b$, $c$, and $d$ respectively and the side of the square is $2.0\text{}\text{cm}$.

Write the expression for electric field at point the center due to all four charges

    $\overrightarrow{E}={\overrightarrow{E}}_a+{\overrightarrow{E}}_b+{\overrightarrow{E}}_c+{\overrightarrow{E}}_d$                                                                                (I)

Here, $\overrightarrow{E}$ is the net electric field, ${\overrightarrow{E}}_a$, ${\overrightarrow{E}}_b$, ${\overrightarrow{E}}_c$, and ${\overrightarrow{E}}_d$ is the electric field due to the charges at the corners $a$, $b$, $c$, and $d$ respectively.

The electrical field due to equal charges at the diagonally opposite corners of the square is equal and opposite. Thus,

    ${\overrightarrow{E}}_b=-{\overrightarrow{E}}_d$                                                                                             (II)

Substitute (II) and (III) in (I)

  $\begin{array}{c}\overrightarrow{E}={\overrightarrow{E}}_a-{\overrightarrow{E}}_d+{\overrightarrow{E}}_c+{\overrightarrow{E}}_d\\ ={\overrightarrow{E}}_a+{\overrightarrow{E}}_c\end{array}$                                                                         (III)

The electric field due to the charge at $a$ is away from $a$ since the charge is positive and the electric field due to charge at $c$ is towards $c$ since the charge is negative.

Substitute for ${\overrightarrow{E}}_a$ and ${\overrightarrow{E}}_c$ in (III)

  $\begin{array}{c}\overrightarrow{E}=\left(\frac{k{q}_a}{r^2}\right)\widehat{r}+\left(\frac{k{q}_c}{r^2}\right)\widehat{r}\\ =\left(\frac{k\left(q_a+q_c\right)}{r^2}\right)\widehat{r}\end{array}$                                                                              (IV)

Here, $k$ is a Coulomb’s constant, $q_a$ is the charge at $a$ and $q_c$ is the charge at $c$, $r$ is the distance between the charge and the center of the square and $\widehat{r}$ is the direction of the electric field i.e. along the line joining the center and the corner $c$.

Write the expression for the distance between the center of the square and its corner.

    $r=\frac{\sqrt{a^2+a^2}}{2}=\frac{a}{\sqrt{2}}$                                                                                    (V)

Here, $a$ is the side of the square.

Substitute (V) in (IV)

  $\overrightarrow{E}=\left(\frac{k\left(q_a+q_c\right)}{{\left(a/\sqrt{2}\right)}^2}\right)\widehat{r}=\left(\frac{2k\left(q_a+q_c\right)}{a^2}\right)\widehat{r}$                                                       (VI)

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}$ for $k$, $+9.0\text{ μC}$ for $q_a$ and $-3.0\text{μC}$ for $q_c$ in (IV) to find $\overrightarrow{E}$

  $\begin{array}{c}\overrightarrow{E}=\frac{2\times 8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(+9.0\text{ μC}-3.0\text{μC}\right)}{{\left(2.0\text{cm}\right)}^2}\widehat{r}\\ =\frac{2\times 8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(6.0\times {10}^{-6}\text{ C}\right)}{{\left(2.0\times {10}^{-2}\text{m}\right)}^2}\widehat{r}\\ =\left(5.4\times {10}^8{\text{NC}}^{\text{-1}}\right)\widehat{r}\end{array}$

Thus, the net electric field is $5.4\times {10}^8{\text{NC}}^{\text{-1}}$ towards the corner $c$.

Write the expression for potential at the center of the square due to the four charges

    $V={\displaystyle \sum _{i=1}^4\frac{k{Q}_i}{r_i}}$                                                                                       (VII)

Here, $V$ is the electric potential, $k$ is the constant, $Q_i$ is the i^th charge and $r_i$ is the distance between the center of the square and the i^th charge.

Since all the charges are equidistant from the center, the above equation reduce to the following

    $V=\left(\frac{k}{r}\right){\displaystyle \sum _{i=1}^4{Q}_i}$                                                                                 (VIII)

Find the sum of all four charges

  $\begin{array}{c}{\displaystyle \sum _{i=1}^4{Q}_i}=\left(+9.0-3(3.0)\right)\text{ μC}\\ \text{=0}\end{array}$

Substitute $0$ for ${\displaystyle \sum _{i=1}^4{Q}_i}$ in (VIII)

$V=0$

Thus, the electric potential is $0$.