Chapter 26, Problem 27PQ

(a)

To determine

The electric potential due to the two charges at the origin.

Answer to Problem 27PQ

The electric potential due to the two charges at the origin is $\underset{\_}{-8.21\times {10}^5\text{V}}$.

Explanation of Solution

The total electric potential at origin is the sum of the potential from individual charges.

Write the expression for total electric potential.

  $\begin{array}{c}V=k_e{\displaystyle \sum _{i=1}^2\frac{q_i}{r_i}}\\ =k_e\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)\end{array}$                                                                                                  (I)

Here, $q_1$ and $q_2$ are charges, $r_1$ and $r_2$ are distances from origin.

Conclusion:

Substitute, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $-6.75\times {10}^{-6}\text{C}$ for $q_1$, $3.20\times {10}^{-6}\text{C}$ for $q_2$, $3.25\times {10}^{-2}\text{m}$ for $r_1$, and $2.75\times {10}^{-2}\text{m}$ for $r_2$ in equation (I).

  $\begin{array}{c}V=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(\frac{-6.75\times {10}^{-6}\text{C}}{3.25\times {10}^{-2}\text{m}}+\frac{3.20\times {10}^{-6}\text{C}}{2.75\times {10}^{-2}\text{m}}\right)\\ =-8.21\times {10}^5\text{V}\end{array}$

Therefore, the electric potential due to the two charges at the origin is $\underset{\_}{-8.21\times {10}^5\text{V}}$.

(b)

To determine

The electric potential due to the two charges at $\left(3.00\text{cm,0}\right)$.

Answer to Problem 27PQ

The electric potential due to the two charges at $\left(3.00\text{cm,0}\right)$ is $\underset{\_}{-6.65\times {10}^5\text{V}}$.

Explanation of Solution

The first $q_A$ is at $\left(0.325\text{cm,0}\right)$ and the point is at $\left(3.00\text{cm,0}\right)$.

Write the expression for the distance between two points if the coordinates are given.

  $r_A=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$                                                                              (II)

For $q_A$, $\left(x_1,y_1\right)$ is $\left(0.325\text{cm,0}\right)$, and $\left(x_2,y_2\right)$ is $\left(3.00\text{cm,0}\right)$, and for $q_B$, $\left(x_1,y_1\right)$ is $\left(0,2.75\text{cm}\right)$, and $\left(x_2,y_2\right)$ is $\left(3.00\text{cm,0}\right)$.

Use equation (I) to find the potential due to two charges at $\left(3.00\text{cm,0}\right)$.

Conclusion:

Substitute, $0.0325\text{m}$ for $y_1$, $0$ for $y_2$, $0$ for $x_1$, and $0.030\text{m}$ for $x_2$ in equation (II) to the find distance from charge $q_A$ to the point in space.

  $\begin{array}{c}{r}_A=\sqrt{{\left(0-0.0325\text{m}\right)}^2+{\left(0.030\text{m}-0\right)}^2}\\ =\sqrt{{\left(-0.0325\text{m}\right)}^2+{\left(0.030\text{m}\right)}^2}\end{array}$

Substitute, $0.0275\text{m}$ for $y_1$, $0$ for $y_2$, $0$ for $x_1$, and $0.030\text{m}$ for $x_2$ in equation (II) to the find distance from charge $q_B$ to the point in space.

  $\begin{array}{c}{r}_B=\sqrt{{\left(0.030\text{m}-0\right)}^2+{\left(0-0.0275\text{m}\right)}^2}\\ =\sqrt{{\left(0.030\text{m}\right)}^2+{\left(-0.0275\text{m}\right)}^2}\end{array}$

Substitute, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $-6.75\times {10}^{-6}\text{C}$ for $q_1$, $3.20\times {10}^{-6}\text{C}$ for $q_2$, $\sqrt{{\left(-0.0325\text{m}\right)}^2+{\left(0.0300\text{m}\right)}^2}$ for $r_A$, and $\sqrt{{\left(0.0300\text{m}\right)}^2+{\left(-0.0275\text{m}\right)}^2}$ for $r_B$ in equation (I).

  $\begin{array}{c}V=\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(\frac{-6.75\times {10}^{-6}\text{C}}{\sqrt{{\left(-0.0325\text{m}\right)}^2+{\left(0.0300\text{m}\right)}^2}}+\frac{3.20\times {10}^{-6}\text{C}}{\sqrt{{\left(0.0300\text{m}\right)}^2+{\left(-0.0275\text{m}\right)}^2}}\right)\\ =-6.65\times {10}^5\text{V}\end{array}$

Therefore, the electric potential due to the two charges at $\left(3.00\text{cm,0}\right)$ is $\underset{\_}{-6.65\times {10}^5\text{V}}$.