Chapter 26, Problem 35PQ

Figure P26.35 shows four particles with identical charges of +5.75 μC arrayed at the vertices of a rectangle of width 25.0 cm and height 55.0 cm. What is the change in the electric potential energy of this system if particles A, B, and C are held in place and particle D is brought from infinity to the position shown in the figure?

FIGURE P26.35

To determine

The change in the electric potential energy of the system if particles $A$ , $B$ and $C$ are held in place and particle $D$ is brought from infinity to the position shown in the figure.

Answer to Problem 35PQ

The change in the electric potential energy of the system if particles $A$ , $B$ and $C$ are held in place and particle $D$ is brought from infinity to the position shown in the figure is $2.22\text{J}$ .

Explanation of Solution

When additional charge is included in the system, the change in potential energy arises due to the effect of charge in existing potentials.

Write the expression for the change in potential energy due to bringing the charge to the point $D$ in the presence of the other three other charges.

  $U_E=q_D{V}_A+q_D{V}_B+q_D{V}_C$                                                                                           (I)

Here, $U_E$ is the change in the potential energy due to bringing additional change at point $D$, $Q_D$ is the charge at $D$ , $V_A$ is the potential due to charge $q_A$ at point $D$ , $V_B$ is the potential due to charge $q_B$ at point $D$ , $V_C$ is the potential due to charge $q_C$ at point $D$.

Write the expression for $V_A$ .

  $V_A=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_A}{r_A}\right)$                                                                                                       (II)

Here, ${\epsilon }_0$ is absolute permittivity and $r_A$ is the distance between points $A$ and $D$.

Write the expression for $V_B$ .

  $V_B=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_B}{r_B}\right)$                                                                                                       (III)

Here, $r_B$ is the distance between points $B$ and $D$.

Write the expression for $V_C$ .

  $V_C=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_C}{r_C}\right)$                                                                                                      (IV)

Here, $r_C$ is the distance between points $C$ and $D$.

Use equations (II), (III) and (IV) in equation (I) to get change in potential energy.

  $U_E=q_D\left(\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_A}{r_A}\right)\right)+q_D\left(\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_B}{r_B}\right)\right)+q_D\left(\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_C}{r_C}\right)\right)$

Since all charges are identical, substitute $q_D$ for $q_A$ , $q_B$ and $q_C$ in above equation.

  $U_E=q_D^2\frac{1}{4\pi {\epsilon }_0}\left(\frac{1}{r_A}+\frac{1}{r_B}+\frac{1}{r_C}\right)$                                                                                  (V)

Write the expression for $r_A$.

  $r_A=\sqrt{A{B}^2+B{D}^2}$                                                                                                  (VI)

Conclusion:

From figureP26.35, distance between point $B$ and $D$ is $55.0\text{cm}$ , distance between point $C$ and $D$ is $25.0\text{cm}$ and distance between point $A$ and $D$ is obtained from Pythagoras Theorem.

Substitute $25.0\text{cm}$ for $AB$ and $55.0\text{cm}$ for $BD$ in equation (VI) to get $r_A$

  $\begin{array}{c}{r}_A=\sqrt{{\left(25.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2+{\left(55.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =0.604\text{m}\end{array}$

Substitute $5.75\text{μC}$ for $q_D$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\frac{1}{4\pi {\epsilon }_0}$ , $55.0\text{cm}$ for $r_B$ and $25.0\text{cm}$ for $r_C$ and $0.604\text{m}$ for $r_A$ in equation (V) to get $U_E$ .

  $\begin{array}{c}{U}_E=\left(5.75\text{μC}\times \frac{1\text{C}}{{10}^6\text{μC}}\right)\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(\frac{1}{0.604\text{m}}+\frac{1}{55.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}}+\frac{1}{25.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}}\right)\\ =2.22\text{J}\end{array}$

Therefore, the change in the electric potential energy of the system if particles $A$ , $B$ and $C$ are held in place and particle $D$ is brought from infinity to the position shown in the figure is $2.22\text{J}$