Chapter 26, Problem 24PQ

Two point charges, q₁ = −2.0 μC and q₂ = 2.0 μC, are placed on the x axis at x = 1.0 m and x = −1.0 m, respectively (Fig. P26.24).

1. a. What are the electric potentials at the points P (0, 1.0 m) and R (2.0 m, 0)?
2. b. Find the work done in moving a 1.0-μC charge from P to R along a straight line joining the two points.
3. c. Is there any path along which the work done in moving the charge from P to R is less than the value from part (b)? Explain.

To determine

The electric potential at $\left(0,1.0\text{m}\right)$, and $\left(2.0\text{m,0}\right)$.

Answer to Problem 24PQ

The electric potential at $P$ is $\underset{\_}{0}$, and electric potential at $R$ is $\underset{\_}{-1.2\times {10}^{-2}\text{V}}$.

Explanation of Solution

Write the expression for electric potential due to two charges.

  $V=\frac{k{q}_1}{r_1}+\frac{k{q}_2}{r_2}$                                                                                                             (I)

Write the expression for distance between two points.

  $r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$                                                                                      (II)

Conclusion:

Consider the figure 1 below. $r_1$ is the distance from $q_1$ to $P$, and $r_2$ is the distance from $q_2$ to $P$

Substitute, $1.0\text{m}$ for $x_2$, $0$ for $x_1$, $0$ for $y_1$, and $1.0\text{m}$ for $y_2$ in equation (II) to obtain distance $r_1$.

  $\begin{array}{c}{r}_1=\sqrt{{\left(-1.0\text{m}-0\right)}^2+{\left(0-1.0\text{m}\right)}^2}\\ =\sqrt{{\left(1\text{m}\right)}^2+{\left(1\text{m}\right)}^2}\\ =\sqrt{2}\end{array}$

Substitute, $-1.0\text{m}$ for $x_2$, $0$ for $x_1$, $0$ for $y_1$, and $1.0\text{m}$ for $y_2$ in equation (II) to obtain distance $r_2$.

  $\begin{array}{c}{r}_2=\sqrt{{\left(1.0\text{m}-0\right)}^2+{\left(0-1.0\text{m}\right)}^2}\\ =\sqrt{{\left(1\text{m}\right)}^2+{\left(1\text{m}\right)}^2}\\ =\sqrt{2}\end{array}$

Substitute, $\sqrt{2}\text{m}$ for $r_1$, and $r_2$, $-2.0\times {10}^{-6}\text{C}$ for $q_1$, and $2.0\times {10}^{-6}\text{C}$ for $q_2$ in equation (I) to obtain the electric potential at $P$.

  $\begin{array}{c}{V}_P=k\left(\frac{-2.0\times {10}^{-6}\text{C}}{\sqrt{2}}+\frac{2.0\times {10}^{-6}\text{C}}{\sqrt{2}}\right)\\ =0\end{array}$

Consider figure 2 given below. $r_1$ is the distance from $q_1$ to $R$, and $r_2$ is the distance from $q_2$ to $R$

Substitute, $2.0\text{m}$ for $x_2$, $1.0\text{m}$ for $x_1$, $0$ for $y_1$, and $0$ for $y_2$ in equation (II) to obtain distance $r_1$.

  $\begin{array}{c}{r}_1=\sqrt{{\left(2.0\text{m}-1.0\text{m}\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{{\left(1\text{m}\right)}^2}\\ =1\text{m}\end{array}$

Substitute, $2.0\text{m}$ for $x_2$, $-1.0\text{m}$ for $x_1$, $0$ for $y_1$, and $0$ for $y_2$ in equation (II) to obtain distance $r_2$.

  $\begin{array}{c}{r}_2=\sqrt{{\left(2.0\text{m}-\left(-1.0\text{m}\right)\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{{\left(9\text{m}\right)}^2}\\ =3\text{m}\end{array}$

Substitute, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $1\text{m}$ for $r_1$, $3.0\text{m}$ for $r_2$, $-2.0\times {10}^{-6}\text{C}$ for $q_1$, and $2.0\times {10}^{-6}\text{C}$ for $q_2$ in equation (I) to obtain the electric potential at $R$.

  $\begin{array}{c}{V}_R=8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\left(\frac{-2.0\times {10}^{-6}\text{C}}{1\text{m}}+\frac{2.0\times {10}^{-6}\text{C}}{3.0\text{m}}\right)\\ =-1.2\times {10}^4\text{V}\end{array}$

Therefore, the electric potential at $P$ is $\underset{\_}{0}$, and electric potential at $R$ is $\underset{\_}{-1.2\times {10}^4\text{V}}$.

To determine

The work done in moving a $1.0\mu \text{C}$ charge from P to R along a straight line joining the two points.

Answer to Problem 24PQ

The work done in moving a $1.0\mu \text{C}$ charge from P to R along a straight line joining the two points is $\underset{\_}{-1.2\times {10}^{-2}\text{J}}$.

Explanation of Solution

The work done will be equal to change in potential energy between two points.

Write the expression for change in electric potential energy.

  $\begin{array}{c}{W}_{P\to R}=\Delta U_E\\ =q\left(V_R-V_P\right)\end{array}$                                                                                                     (I)

Here, $q$ is the charge of the particle.

Conclusion:

Substitute, $1.0\times {10}^{-6}\text{C}$ for $q$, $-1.2\times {10}^4\text{V}$ for $V_R$, and $0$ for $V_P$ in equation (I).

  $\begin{array}{c}{W}_{p\to R}=1.0\times {10}^{-6}\text{C}\left(-1.2\times {10}^4\text{V}-0\right)\\ =-1.2\times {10}^{-2}\text{J}\end{array}$

Therefore, the work done in moving a $1.0\mu \text{C}$ charge from P to R along a straight line joining the two points is $\underset{\_}{-1.2\times {10}^{-2}\text{J}}$.

To determine

The path along which the work done is less than the value obtained in part (b).

Answer to Problem 24PQ

No, there is no other path through which the charge can move so that the work done is less than the value obtained part (b).

Explanation of Solution

Write the expression for work done in terms of change in potential energy.

  $\begin{array}{c}W=\Delta U_E\\ =q\Delta V\end{array}$

The work done is depends on charge and the change in potential, and does not depends on the path followed by the particle.

Hence, there is no other path in which work done is less than $-1.2\times {10}^{-2}\text{J}$.

Conclusion:

Therefore, the work done is not depends on the path followed by the particle, and only depends on the change in potential. There is no other path with work done less than $-1.2\times {10}^{-2}\text{J}$.