Chapter 17, Problem 31P

(a)

To determine

The electric potential at the midpoint of two point charges.

Answer to Problem 31P

The electric potential at the midpoint of two point charges is $7.35\times {10}^4\text{V}$.

Explanation of Solution

The charges are $35.0\text{ nC}$ and $55.0\text{ nC}$. The first charge is placed at the origin and the second charge is placed at a distance $2.20\text{cm}$ from the origin.

Write the expression for potential at point due to two charges

    $V_a=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)$                                                                                         (I)

Here, $V_a$ is the electric potential at the midpoint of the two charges, $k$ is the Coulomb’s constant, $q_1$, $q_2$ are the charges and $r_1$, $r_2$  are their corresponding distance between the charges and the midpoint of the two charges.

Write the expression for midpoint

  $M=\frac{x_1+x_2}{2}$                                                                                                   (II)

Here, $M$ is the midpoint of the two charges, $x_1$ is the position of the first charge and $x_2$ is the position of the second charge.

Substitute $0$ for $x_1$ and $2.2\text{cm}$ for $x_2$ in (II) to find $M$

  $\begin{array}{c}M=\frac{\left(0+2.20\right)\text{cm}}{2}\\ =1.1\text{ cm}\end{array}$

The two charges are located on either side of the midpoint. Thus the distances $r_1$ and $r_2$ are same.

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}$ for $k$, $35.0\text{ nC}$ for $q_1$, $55.0\text{ nC}$ for $q_2$, $1.1\text{ cm}$ for $r_1$ and $1.1\text{ cm}$ for $r_2$ in (I) to find $V_a$. Here, $V_a$ is the potential at the midpoint.

  $\begin{array}{c}{V}_a=8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(\frac{35.0\text{ nC}}{1.1\text{ cm}}+\frac{55.0\text{ nC}}{1.1\text{ cm}}\right)\\ =8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(\frac{35.0\text{ C}}{1.1\text{ m}}+\frac{35.0\text{ C}}{1.1\text{ m}}\right)\times \frac{{10}^{-9}}{{10}^{-2}}\\ =7.35\times {10}^4\text{V}\end{array}$

Thus, the electric potential at the midpoint is $7.35\times {10}^4\text{V}$.

(b)

To determine

The electric potential at a point $+3.40\text{cm}$ from the origin.

Answer to Problem 31P

The electric potential at a distance $+3.40\text{cm}$ from the origin is $5.04\times {10}^4\text{V}$.

Explanation of Solution

The first charge is at a distance $+3.40\text{cm}$ from the point and the second charge is at a distance $3.40\text{cm}-2.20\text{cm}$ from the point. (Refer figure 1)

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}$ for $k$, $35.0\text{ nC}$ for $q_1$, $55.0\text{ nC}$ for $q_2$, $+3.40\text{cm}$ for $r_1$ and $3.40\text{cm}-2.20\text{cm}$ for $r_2$ in (I) to find $V_b$. Here, $V_a$ is the potential at a distance $+3.40\text{cm}$.

  $\begin{array}{c}{V}_b=8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(\frac{35.0\text{ nC}}{3.40\text{ cm}}+\frac{55.0\text{ nC}}{3.40\text{cm}-2.20\text{cm}}\right)\\ =8.988\times {10}^9{\text{Nm}}^{\text{2}}{\text{C}}^{-\text{2}}\left(\frac{35.0\text{ C}}{3.40\text{ m}}+\frac{35.0\text{ C}}{1.2\text{ m}}\right)\times \frac{{10}^{-9}}{{10}^{-2}}\\ =5.04\times {10}^4\text{V}\end{array}$

Thus, the electric potential at a distance $+3.40\text{cm}$ from the origin is $5.04\times {10}^4\text{V}$.

(c)

To determine

The work done by an external force in moving a charge at a distance $+3.40\text{cm}$ from the origin

to the midpoint of the two charges.

Answer to Problem 31P

The work done is $1.04\text{mJ}$.

Explanation of Solution

The charge being moved is $45.0\text{nC}$.

Write the expression for work done by an external agent in moving a charge

  $W=\Delta U$                                                                                                  (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 $\Delta U$

  $\Delta U=q\left(V_{\text{f}}-V_{\text{i}}\right)$                                                                                         (IV)

Here, $q$ is the charge being moved $V_{\text{i}}$ is the potential at a distance $+3.40\text{cm}$ from the origin and  $V_{\text{f}}$ is the potential at the midpoint of the two charges.

Equate (IV) and (III)

  $W=q\left(V_{\text{f}}-V_{\text{i}}\right)$                                                                                             (V)

Substitute  $45.0\text{nC}$ for $q$, $5.04\times {10}^4\text{V}$ for $V_{\text{i}}$ and $7.35\times {10}^4\text{V}$ for $V_{\text{f}}$ in (V) to find $W$

    $\begin{array}{c}W=45.0\text{nC}\left[\left(7.35\times {10}^4\text{V}\right)-\left(5.04\times {10}^4\text{V}\right)\right]\\ =+2.00\times {10}^{-9}\text{C}\left(-899\text{V}-1348\text{V}\right)\\ =1.04\times {10}^{-3}\text{J}\\ =1.04\text{mJ}\end{array}$

Thus, the work done is $1.04\text{mJ}$.