Chapter 17, Problem 27P

(a)

To determine

The electric potential at a point due to two charges.

Answer to Problem 27P

The electric potential at points $a$ and $b$ are $270\text{V}$ and $-160\text{ V}$.

Explanation of Solution

The side of the square is $1.0\text{m}$ and the charges are $+4.2\text{ nC}$ and $-6.4\text{nC}$.

Write the expression for potential at the point due to two point charges.

    $V=k\left(\frac{Q_{\text{+}}}{r_{\text{+}}}+\frac{Q_{-}}{r_{-}}\right)$                                                              (I)

Here, $V$ is the electric potential, $k$ is the Coulomb’s constant, $Q_{+}$ and $Q_{-}$ are the charges whereas $r_{+}$ and $r_{-}$  are the corresponding distance between the charges and a point in the figure.

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}$ for $k$, $+4.2\text{ nC}$ for $Q_{+}$, $-6.4\text{nC}$ for $Q_{-}$, $6.0\text{}\text{cm}$ for $r_{+}$ and $15.9\text{cm}$ for $r_{-}$ in (I) to find $V_a$. Here, $V_a$ is the potential at the point $a$.

  $\begin{array}{c}{V}_a=8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(\frac{+4.2\text{ nC}}{6.0\text{}\text{cm}}+\frac{-6.4\text{nC}}{15.9\text{}\text{cm}}\right)\\ =8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(\frac{+4.2\text{ C}}{6.0\text{}\text{m}}+\frac{-6.4\text{C}}{6.0\text{}\text{m}}\right)\times \frac{{10}^{-9}}{{10}^{-2}}\\ =270\text{V}\end{array}$

Thus, the electric potential at $a$ is $270\text{V}$.

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}$ for $k$, $+4.2\text{ nC}$ for $Q_{+}$, $-6.4\text{nC}$ for $Q_{-}$, $12.0\text{}\text{cm}$ for $r_{+}$ and $12.0\text{cm}$ for $r_{-}$ in (I) to find $V_b$. Here, $V_b$ is the potential at the point $b$.

  $\begin{array}{c}{V}_b=8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(\frac{+4.2\text{ nC}}{12.0\text{}\text{cm}}+\frac{-6.4\text{nC}}{12.0\text{}\text{cm}}\right)\\ =8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(\frac{+4.2\text{ C}}{12.0\text{}\text{m}}+\frac{-6.4\text{C}}{12.0\text{}\text{m}}\right)\times \frac{{10}^{-9}}{{10}^{-2}}\\ =-160\text{V}\end{array}$

Thus, the electric potential at $b$ is $-160\text{V}$.

(b)

To determine

The potential difference in travelling from $b$ to $a$.

Answer to Problem 27P

The potential difference is $430\text{V}$.

Explanation of Solution

Write the expression for potential difference

    $\Delta V=V_{\text{f}}-V_{\text{i}}$                                                               (II)

Here, $\Delta V$ is the potential difference, $V_{\text{f}}$ is the potential at the final point and $V_{\text{i}}$ is the potential at the initial point.

Substitute $270\text{V}$ for $V_{\text{f}}$ and $-160\text{V}$ for $V_{\text{i}}$ in (II) to find $\Delta V$

  $\begin{array}{c}\Delta V=270\text{V}-(-160\text{V})\\ =430\text{ V}\end{array}$

Thus, the potential difference is $430\text{V}$.

(c)

To determine

The work done by an external force in moving a charge from $b$ to $a$.

Answer to Problem 27P

The work done is $6.5\times {10}^{-7}\text{J}$.

Explanation of Solution

The charge being moved is $+1.50\text{}\text{ nC}$.

Write the expression for work done in moving a charge from $b$ to $a$.

  $W=\Delta U$                                                  (III)

Here, $W$ is the work done by the external force and $\Delta U$ is the change in electric potential in moving the charge from $b$ to $a$.

Write the expression for $\Delta U$

  $\Delta U=q\Delta V$                                                   (IV)

Here, $q$ is the charge being moved and $\Delta V$ is the potential difference between point $a$ and $b$.

Substitute (IV) in (III)

  $W=q\Delta V$                                                  (V)

Substitute $+1.50\text{}\text{ nC}$ for $q$ and $430\text{ V}$ for $\Delta V$ in (V) to find $W$

    $\begin{array}{c}W=\left(+1.50\text{}\text{ nC}\right)\left(430\text{V}\right)\\ =\left(+1.50\times {10}^{-9}\text{}\text{C}\right)\left(430\text{V}\right)\\ =6.5\times {10}^{-7}\text{J}\end{array}$

Thus, the work done is $6.5\times {10}^{-7}\text{J}$.