Chapter 23, Problem 27P

(a)

To determine

The electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_2=q_3=+2\text{μC}$ .

Answer to Problem 27P

The electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_2=q_3=+2\text{μC}$ is $12.9\text{kV}$ .

Explanation of Solution

Given:

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

Formula used:

The expression for the potential at the point whose coordinates is $\left(0,3\right)$ ,

  $\begin{array}{c}V=\frac{kq}{r_1}+\frac{kq}{r_2}+\frac{kq}{r_3}\\ =kq\left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\right)\end{array}$

Calculation:

  

Figure (1)

The distance to the point $P$ from the charge $q_1$ is $r_1=3\text{m}$

The distance to the point $P$ from the charge $q_2$ is,

  $\begin{array}{c}{r}_2=\sqrt{{\left(3\text{m}\right)}^2+{\left(3\text{m}\right)}^2}\\ =4.243\text{m}\end{array}$

The distance to the point $P$ from the charge $q_3$ is,

  $\begin{array}{c}{r}_3=\sqrt{{\left(3\text{m}\right)}^2+{\left(6\text{m}\right)}^2}\\ =6.71\text{m}\end{array}$

The potential at the point is calculated as,

  $\begin{array}{c}V=kq\left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\times \left(2\text{μC}\right)\times \left(\frac{1}{3\text{m}}+\frac{1}{4.243\text{m}}+\frac{1}{6.71\text{m}}\right)\\ =12.9\text{kV}\end{array}$

Conclusion:

Therefore, the electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_2=q_3=+2\text{μC}$ is $12.9\text{kV}$ .

(b)

To determine

The electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_2=+2\text{μC}$ and $q_3=-2.00\text{μC}$ .

Answer to Problem 27P

The electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_2=+2\text{μC}$ and $q_3=-2.00\text{μC}$ is $7.55\text{kV}$ .

Explanation of Solution

Given:

The charge on first and second particle is $q_1=q_2=+2.00\text{μC}$

The charge on third particle is $q_3=-2.00\text{μC}$

Formula used:

The expression for the potential at the point whose coordinates is $\left(0,3\right)$ ,

  $\begin{array}{c}V=\frac{kq}{r_1}+\frac{kq}{r_2}+\frac{kq}{r_3}\\ =k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\right)\end{array}$

Calculation:

The distance to the point $P$ from the charge $q_1$ is $r_1=3\text{m}$

The distance to the point $P$ from the charge $q_2$ is,

  $\begin{array}{c}{r}_2=\sqrt{{\left(3\text{m}\right)}^2+{\left(3\text{m}\right)}^2}\\ =4.243\text{m}\end{array}$

The distance to the point $P$ from the charge $q_3$ is,

  $\begin{array}{c}{r}_3=\sqrt{{\left(3\text{m}\right)}^2+{\left(6\text{m}\right)}^2}\\ =6.71\text{m}\end{array}$

The potential at the point is calculated as,

  $\begin{array}{c}V=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\times \left[\frac{2\text{μC}}{3\text{m}}+\frac{2\text{μC}}{4.243\text{m}}+\frac{-\left(2\text{μC}\right)}{6.71\text{m}}\right]\\ =7.55\text{kV}\end{array}$

Conclusion:

Therefore, the electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_2=+2\text{μC}$ and $q_3=-2.00\text{μC}$ is $7.55\text{kV}$ .

(c)

To determine

The electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_2=+2\text{μC}$ and $q_3=-2.00\text{μC}$ .

Answer to Problem 27P

The electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_3=+2\text{μC}$ and $q_2=-2.00\text{μC}$ is $4.43\text{kV}$ .

Explanation of Solution

Given:

The charge on first and third particle is $q_1=q_3=+2.00\text{μC}$

The charge on second particle is $q_2=-2.00\text{μC}$

Formula used:

The expression for the potential at the point whose coordinates is $\left(0,3\right)$ ,

  $\begin{array}{c}V=\frac{kq}{r_1}+\frac{kq}{r_2}+\frac{kq}{r_3}\\ =k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\right)\end{array}$

Calculation:

The distance to the point $P$ from the charge $q_1$ is $r_1=3\text{m}$

The distance to the point $P$ from the charge $q_2$ is,

  $\begin{array}{c}{r}_2=\sqrt{{\left(3\text{m}\right)}^2+{\left(3\text{m}\right)}^2}\\ =4.243\text{m}\end{array}$

The distance to the point $P$ from the charge $q_3$ is,

  $\begin{array}{c}{r}_3=\sqrt{{\left(3\text{m}\right)}^2+{\left(6\text{m}\right)}^2}\\ =6.71\text{m}\end{array}$

The potential at the point is calculated as,

  $\begin{array}{c}V=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\right)\\ =\left(8.988\times {10}^9{\text{Nm}}^2/{\text{C}}^2\right)\times \left[\frac{2\text{μC}}{3\text{m}}+\frac{-2\text{μC}}{4.243\text{m}}+\frac{\left(2\text{μC}\right)}{6.71\text{m}}\right]\\ =4.43\text{kV}\end{array}$

Conclusion:

Therefore, The electric potential at the point on $y$ -axis at $y=3.00\text{m}$ if $q_1=q_3=+2\text{μC}$ and $q_2=-2.00\text{μC}$ is $4.43\text{kV}$ .