Chapter 17, Problem 90QAP

To determine

(a)

The electric potential energy of the charge distribution

Answer to Problem 90QAP

Potential energy of the charge distribution is $=\frac{3K{q}^2}{a}$

Explanation of Solution

Given:

Side of a rhombus is a
Short diagonal is length a and long diagonal is of length b
Calculation:

Potential energy of the charge distribution is
  $\begin{array}{l}{U}_{total}=U_{q_1{q}_2}+U_{q_2{q}_3}+U_{q_1{q}_3}\\ =\frac{K{q}_1{q}_2}{a}+\frac{K{q}_2{q}_3}{a}+\frac{K{q}_1{q}_3}{a}\\ =\frac{K}{a}\left[q_1{q}_2+q_2{q}_3+q_1{q}_3\right]\\ =\frac{K}{a}\left[q^2+q^2+q^2\right]\\ =\frac{3K{q}^2}{a}\end{array}$

Conclusion:

Potential energy of the charge distribution is $=\frac{3K{q}^2}{a}$

To determine

(b)

The potential energy at the vacant end of the rhombus

Answer to Problem 90QAP

The work required to bring the fourth particle $=\frac{K{q}^2}{a}\left(\frac{1}{\sqrt{3}}+2\right)$

Explanation of Solution

The potential energy at the vacant end of the rhombus is zero

To determine

(c)

To calculate the work required to bring the fourth particle

Answer to Problem 90QAP

The work required to bring the fourth particle $=\frac{K{q}^2}{a}\left(\frac{1}{\sqrt{3}}+2\right)$

Explanation of Solution

Given:

Side of a rhombus is a
Short diagonal is length a and long diagonal is of length b
Calculation:

  $\begin{array}{l}W=\Delta U\\ =U_f-U_i\\ =U_f-0\\ =U_{q_1{q}_4}+U_{q_2{q}_4}+U_{q_3{q}_4}\\ =\frac{K{q}_1{q}_2}{b}+\frac{K{q}_2{q}_3}{a}+\frac{K{q}_1{q}_3}{a}\\ =\frac{K{q}_1{q}_2}{\sqrt{3a^2}}+\frac{K{q}_2{q}_3}{a}+\frac{K{q}_1{q}_3}{a}\\ =\frac{K{q}^2}{\sqrt{3}a}+\frac{2K{q}^2}{a}\\ =\frac{K{q}^2}{a}\left(\frac{1}{\sqrt{3}}+2\right)\end{array}$

Conclusion:

The work required to bring the fourth particle $=\frac{K{q}^2}{a}\left(\frac{1}{\sqrt{3}}+2\right)$

To determine

(d)

The total potential energy of four charges

Answer to Problem 90QAP

The total potential energy of four charges $=\frac{K{q}^2}{a}\left(\frac{1}{\sqrt{3}}+5\right)$

Explanation of Solution

Given:

Side of a rhombus is a
Short diagonal is length a and long diagonal is of length b
Calculation:

  $\begin{array}{l}{U}_{total}=U_{q_1{q}_2}+U_{q_2{q}_3}+U_{q_1{q}_3}+U_{q_1{q}_4}+U_{q_2{q}_4}+U_{q_3{q}_4}\\ =\frac{3K{q}^2}{a}+\frac{K{q}^2}{a}\left(\frac{1}{\sqrt{3}}+2\right)\\ =\frac{K{q}^2}{a}\left(\frac{1}{\sqrt{3}}+5\right)\end{array}$

Conclusion:

The total potential energy of four charges $=\frac{K{q}^2}{a}\left(\frac{1}{\sqrt{3}}+5\right)$