Chapter 23, Problem 68P

(a)

To determine

The electrostatic potential energy of the system when all of the charges are negative.

Answer to Problem 68P

The electrostatic potential energy of the system when all of the charges are negative is $48.7\text{mJ}$ .

Explanation of Solution

Given Data:

The charge on each particle is $q=2.00\text{μC}$ .

The side of a square is $a=4\text{m}$ .

Formula used:

The expression for the work required to assemble the system of charges which is equal to the potential energy of the system is given as,

  $U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_3{q}_4}{r_{3,4}}+\frac{q_2{q}_4}{r_{2,4}}+\frac{q_2{q}_3}{r_{2,3}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_1{q}_4}{r_{1,4}}\right)$

Calculation:

  

Figure (1)

The electrostatic potential energy of the system is calculated as,

  $\begin{array}{c}U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_3{q}_4}{r_{3,4}}+\frac{q_2{q}_4}{r_{2,4}}+\frac{q_2{q}_3}{r_{2,3}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_1{q}_4}{r_{1,4}}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(-2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}+\frac{\left(-2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}+\\ \frac{\left(-2\text{μC}\right)\left(-2\text{μC}\right)}{4\sqrt{2}\text{m}}+\frac{\left(-2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}+\\ \frac{\left(-2\text{μC}\right)\left(-2\text{μC}\right)}{4\sqrt{2}\text{m}}+\frac{\left(-2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}\end{array}\right)\\ =48.7\text{mJ}\end{array}$

Conclusion:

Therefore, the electrostatic potential energy of the system when all of the charges are negative is $48.7\text{mJ}$ .

(b)

To determine

The electrostatic potential energy of the system when three of the charges are positive and one of the charge is negative.

Answer to Problem 68P

The electrostatic potential energy of the system when three of the charges are positive and one of the charge is negative is zero.

Explanation of Solution

Given Data:

The charge on three particles is $q_1=q_2=q_3=2\text{μC}$

The charge on fourth particle is $q_4=-2\text{μC}$

The side of a square is $a=4\text{m}$ .

Formula used:

The expression for the work required to assemble the system of charges which is equal to the potential energy of the system is given as,

  $U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_3{q}_4}{r_{3,4}}+\frac{q_2{q}_4}{r_{2,4}}+\frac{q_2{q}_3}{r_{2,3}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_1{q}_4}{r_{1,4}}\right)$

Calculation:

The electrostatic potential energy of the system is calculated as,

  $\begin{array}{c}U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_3{q}_4}{r_{3,4}}+\frac{q_2{q}_4}{r_{2,4}}+\frac{q_2{q}_3}{r_{2,3}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_1{q}_4}{r_{1,4}}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(2\text{μC}\right)\left(2\text{μC}\right)}{4\text{m}}+\frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}+\\ \frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\sqrt{2}\text{m}}+\frac{\left(2\text{μC}\right)\left(2\text{μC}\right)}{4\text{m}}+\\ \frac{\left(2\text{μC}\right)\left(2\text{μC}\right)}{4\sqrt{2}\text{m}}+\frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}\end{array}\right)\\ =0\text{mJ}\end{array}$

Conclusion:

Therefore the electrostatic potential energy of the system when three of the charges are positive and one of the charge is negative is zero.

(c)

To determine

The electrostatic potential energy of the system when two adjacent corners are positive and other two are negative

Answer to Problem 68P

The electrostatic potential energy of the system when two adjacent corners are positive and other two are negative is $-12.7\text{mJ}$

Explanation of Solution

Given Data:

The charge on first particle is $q_1=2\text{μC}$

The charge on second particle is $q_2=2\text{μC}$

The charge on third particle is $q_3=2\text{μC}$

The charge on fourth particle is $q_4=-2\text{μC}$

The side of a square is $a=4\text{m}$ .

Formula used:

The expression for the work required to assemble the system of charges which is equal to the potential energy of the system is given as,

  $U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_3{q}_4}{r_{3,4}}+\frac{q_2{q}_4}{r_{2,4}}+\frac{q_2{q}_3}{r_{2,3}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_1{q}_4}{r_{1,4}}\right)$

Calculation:

The electrostatic potential energy of the system is calculated as,

  $\begin{array}{c}U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_3{q}_4}{r_{3,4}}+\frac{q_2{q}_4}{r_{2,4}}+\frac{q_2{q}_3}{r_{2,3}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_1{q}_4}{r_{1,4}}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(2\text{μC}\right)\left(2\text{μC}\right)}{4\text{m}}+\frac{\left(-2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}+\\ \frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\sqrt{2}\text{m}}+\frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}+\\ \frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\sqrt{2}\text{m}}+\frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}\end{array}\right)\\ =-12.7\text{mJ}\end{array}$

Conclusion:

Therefore the electrostatic potential energy of the system when two adjacent corners are positive and other two are negative is $-12.7\text{mJ}$

(d)

To determine

The electrostatic potential energy of the system when the charges at two opposite corners are positive and other two are negative.

Answer to Problem 68P

The electrostatic potential energy of the system when the charges at two opposite corners are positive and other two are negative is $-23.2\text{mJ}$ .

Explanation of Solution

Given Data:

The charge on first particle is $q_1=2\text{μC}$

The charge on second particle is $q_2=-2\text{μC}$

The charge on third particle is $q_3=2\text{μC}$

The charge on fourth particle is $q_4=-2\text{μC}$

The side of a square is $a=4\text{m}$ .

Formula used:

The expression for the work required to assemble the system of charges which is equal to the potential energy of the system is given as,

  $U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_3{q}_4}{r_{3,4}}+\frac{q_2{q}_4}{r_{2,4}}+\frac{q_2{q}_3}{r_{2,3}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_1{q}_4}{r_{1,4}}\right)$

Calculation:

The electrostatic potential energy of the system is calculated as,

  $\begin{array}{c}U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_3{q}_4}{r_{3,4}}+\frac{q_2{q}_4}{r_{2,4}}+\frac{q_2{q}_3}{r_{2,3}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_1{q}_4}{r_{1,4}}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}+\frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}+\\ \frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\sqrt{2}\text{m}}+\frac{\left(-2\text{μC}\right)\left(2\text{μC}\right)}{4\text{m}}+\\ \frac{\left(2\text{μC}\right)\left(2\text{μC}\right)}{4\sqrt{2}\text{m}}+\frac{\left(2\text{μC}\right)\left(-2\text{μC}\right)}{4\text{m}}\end{array}\right)\\ =-23.2\text{mJ}\end{array}$

Conclusion:

Therefore, the electrostatic potential energy of the system when the charges at two opposite corners are positive and other two are negative is $-23.2\text{mJ}$ .