Chapter 16, Problem 20P

To determine

To Find: The third position of charge

Answer to Problem 20P

The third charge must be placed at 118 cm from the charge $q_1$ .

Explanation of Solution

Given:

Two charges are:

  $+4.75\mu \text{C}$

  $-3.55\mu \text{C}$

Formula used:

Electrostatic force between two charges having magnitudes $q_1,q_2$ , having distance $r$ can be determined by Coulomb’s law as,

  $F=\frac{k{q}_1{q}_2}{r^2}$

Here, $k$ is electrostatic constant $\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)$ .

Calculation:

Below sketch shows positions of the charges.

  

Conversion of units of given two charges from micro coulombs to coulombs,

  $\begin{array}{l}{q}_1=+4.75\mu \text{C}\\ =(+4.75\mu \text{C})\left(\frac{{10}^{-6}\text{C}}{\mu \text{C}}\right)\\ =+4.75\times {10}^{-6}\text{C}\\ q_2=-3.55\mu \text{C}\\ =(-3.55\mu \text{C})\left(\frac{{10}^{-6}\text{C}}{\mu \text{C}}\right)\\ =-3.55\times {10}^{-6}\text{C}\end{array}$

For no net force, electrostatic force between the charges $q,q_1$ should equal to electrostatic force between $q,q_2$ mathematically,

  $\begin{array}{l}\frac{kq{q}_1}{{(r+x)}^2}=-\frac{kq{q}_2}{x^2}\\ \frac{q_1}{{(r+x)}^2}=-\frac{q_2}{x^2}\\ \frac{4.75\times {10}^{-6}\text{C}}{{(18.5\text{cm}+x)}^2}=\frac{3.55\times {10}^{-6}\text{C}}{x^2}\\ \frac{4.75}{3.55}={\left(\frac{18.5+x}{x}\right)}^2\\ 1.157=\left(\frac{18.5+x}{x}\right)\\ x=118\text{cm}\end{array}$

Conclusion:

Therefore, to experience no net force, the third charge must be placed at 118 cm from the charge.