Chapter 20.2, Problem 11PP

To determine

To Draw: Diagram of the situation.

To Find: The net force on A.

Explanation of Solution

Given:

Charge on A, $q_A=+6.00\text{μC}$

Charge on B, $q_B=+3.00\text{μC}$

Charge on C, $q_C=+1.5\text{μC}$

Distance between charge A and B, $d_{AB}=4.0\text{cm}\text{=}\text{0}\text{.04 m}$

Distance between charge A and C, $d_{AC}=3.0\text{cm =0}\text{.03 m}$

Formula Used:

Electrostatic force between two charges:

  $F_{12}=k\frac{q_1{q}_2}{d^2}$

Here, k is the Coulomb’s constant, $q_1$ and $q_2$ are the charges and dis the distance between them.

Calculations:

  

Force between A and B:

$\begin{array}{c}{F}_{AB}=k\frac{q_A{q}_B}{d_{AB}}\\ =\left(9\times {10}^9\right)\frac{\left(6.0\times {10}^{-6}\right)\left(3\times {10}^{-6}\right)}{{0.04}^2}\\ =1.0\times {10}^2\text{N}\end{array}$

Force between A and C:

$\begin{array}{c}{F}_{AC}=k\frac{q_A{q}_C}{d_{AC}}\\ \hspace{1em}\hspace{1em}=\left(9\times {10}^9\right)\frac{\left(6.0\times {10}^{-6}\right)\left(1.5\times {10}^{-6}\right)}{{0.03}^2}\\ =9.0\times {10}^1\text{N}\end{array}$

Both the forces repel. The net force on A due B and C:

  $\begin{array}{c}{F}_{net}=\sqrt{F_{AB}{}^2+F_{Ac}{}^2}\\ =\sqrt{{\left(1.0\times {10}^2\right)}^2+{\left(9.0\times {10}^1\right)}^2}\\ =130\text{N}\end{array}$

Angle above the negative horizontal axis:

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{F_{AC}}{F_{AB}}\right)\\ ={\tan }^{-1}\left(\frac{9.0\times {10}^1}{1.0\times {10}^2}\right)\\ =42°\end{array}$

Angle from the positive horizontal axis: $180°-42°=138°$ .