Chapter 23, Problem 66P

(a)

To determine

The electrostatic potential energy of the system of the charge.

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is $190\text{mJ}$ .

Explanation of Solution

Given Data:

The charge on first particle is $q_1=+4.20\text{μC}$

The charge on second particle is $q_2=+4.20\text{μC}$

The charge on third particle is $q_3=+4.20\text{μC}$

Formula used:

The expression for the work required to assemble the system of charges is given as,

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

Here, the radius of the particle is given as,

  $\begin{array}{l}{r}_{1,2}=2.50\text{m}\\ {\text{r}}_{2,3}=2.50\text{m}\\ {\text{r}}_{1,3}=2.50\text{m}\end{array}$

Calculation:

The work required to assemble the system of charges is calculated as,

  $\begin{array}{c}U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_2{q}_3}{r_{2,3}}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(4.20\text{μC}\right)\left(4.20\text{μC}\right)}{2.50\text{m}}\\ +\frac{\left(4.20\text{μC}\right)\left(4.20\text{μC}\right)}{2.50\text{m}}\\ +\frac{\left(4.20\text{μC}\right)\left(4.20\text{μC}\right)}{2.50\text{m}}\end{array}\right)\\ =190\text{mJ}\end{array}$

Conclusion:

Therefore, the electrostatic potential energy of the system of the charge is $190\text{mJ}$ .

(b)

To determine

The electrostatic potential energy of the system of the charge.

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is $-63.4\text{mJ}$ .

Explanation of Solution

Given Data:

The charge on first particle is $q_1=+4.20\text{μC}$

The charge on second particle is $q_2=+4.20\text{μC}$

The charge on third particle is $q_3=-4.20\text{μC}$

Formula used:

The expression for the work required to assemble the system of charges is given as,

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

Here, the radius of the particle is given as,

  $\begin{array}{l}{r}_{1,2}=2.50\text{m}\\ {\text{r}}_{2,3}=2.50\text{m}\\ {\text{r}}_{1,3}=2.50\text{m}\end{array}$

Calculation:

The work required to assemble the system of charges is calculated as,

  $\begin{array}{c}U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_2{q}_3}{r_{2,3}}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(4.20\text{μC}\right)\left(4.20\text{μC}\right)}{2.50\text{m}}\\ +\frac{\left(4.20\text{μC}\right)\left(-4.20\text{μC}\right)}{2.50\text{m}}\\ +\frac{\left(4.20\text{μC}\right)\left(-4.20\text{μC}\right)}{2.50\text{m}}\end{array}\right)\\ =-63.4\text{mJ}\end{array}$

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is $-63.4\text{mJ}$ .

(c)

To determine

The electrostatic potential energy of the system of the charge.

Answer to Problem 66P

The electrostatic potential energy of the system of the charge is $-63.4\text{mJ}$ .

Explanation of Solution

Given Data:

The charge on first particle is $q_1=-4.20\text{μC}$

The charge on second particle is $q_2=-4.20\text{μC}$

The charge on third particle is $q_3=+4.20\text{μC}$

Formula used:

The expression for the work required to assemble the system of charges is given as,

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

Here, the radius of the particle is given as,

  $\begin{array}{l}{r}_{1,2}=2.50\text{m}\\ {\text{r}}_{2,3}=2.50\text{m}\\ {\text{r}}_{1,3}=2.50\text{m}\end{array}$

Calculation:

The work required to assemble the system of charges is calculated as,

  $\begin{array}{c}U=k\left(\frac{q_1{q}_2}{r_{1,2}}+\frac{q_1{q}_3}{r_{1,3}}+\frac{q_2{q}_3}{r_{2,3}}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(\begin{array}{l}\frac{\left(-4.20\text{μC}\right)\left(-4.20\text{μC}\right)}{2.50\text{m}}\\ +\frac{\left(-4.20\text{μC}\right)\left(4.20\text{μC}\right)}{2.50\text{m}}\\ +\frac{\left(-4.20\text{μC}\right)\left(4.20\text{μC}\right)}{2.50\text{m}}\end{array}\right)\\ =-63.4\text{mJ}\end{array}$

Conclusion:

Therefore the electrostatic potential energy of the system of the charge is $-63.4\text{mJ}$ .