Chapter 16, Problem 53QAP

To determine

The x and y component of the net electric force that acts on the third charge Q₃.

Answer to Problem 53QAP

The x -component of the net electric force is $F_{xon3}=-90.6192N$ and y -component is $F_{yon3}=14.02N$.

Explanation of Solution

Given:

Three charges are located as shown in figure

  $\begin{array}{l}{Q}_1=+7.00\mu \text{C}\\ Q_2=-4.00\mu \text{C}\\ Q_3=+6.00\mu \text{C}\end{array}$

  $\begin{array}{l}{x}_{21}^{}=0.03m\\ x_{23}^{}=0.04m\end{array}$

  

Formula used:

From Coulomb's law

  $\begin{array}{l}{F}_{q_1\text{ on }{q}_2}=\frac{k|q_1\Vert q_2|}{r^2}\\ \text{where},\\ r\text{=}\text{seperation}\text{between}\text{charge}\text{1}\text{and}\text{charge}\text{2}\end{array}$

Calculation:

  $\begin{array}{l}{x}_{13}^2=x_{21}^2+x_{23}^2\\ x_{13}^2={\left(0.03m\right)}^2+{\left(0.04m\right)}^2\\ x_{13}^2={\left(0.05m\right)}^2\end{array}$

Force on charge Q₃ due to charge Q₁.

  $\begin{array}{l}{F}_{1on3}=\frac{k\left|q_1\right|\left|q_3\right|}{x_{13}^2}\\ F_{1on3}=\left(8.99\times {10}^9\frac{N.m^2}{C^2}\right)\frac{\left|7\times {10}^{-6}C\right|\left|6\times {10}^{-6}C\right|}{{\left(0.05m\right)}^2}\\ F_{1on3}=151.032\frac{N}{C}\end{array}$

Charge 1 and charge 3 have the same sign, the repulsive force on the charge 3 due to charge 1 have direction in the third quadrant. So, $F_{1on3}=-151.032N$.

Force on charge Q₃ due to charge Q₂.

  $\begin{array}{l}{F}_{2on3}=\frac{k\left|q_2\right|\left|q_3\right|}{x_{23}^2}\\ F_{2on3}=\left(8.99\times {10}^9\frac{N.m^2}{C^2}\right)\frac{\left|-4\times {10}^{-6}C\right|\left|6\times {10}^{-6}C\right|}{{\left(0.04m\right)}^2}\\ F_{2on3}=134.85\frac{N}{C}\end{array}$

x component of net force on charge Q₃ :

  $\begin{array}{l}{F}_{xon3}=F_{1on3}\cos \theta \\ F_{xon3}=\left(-151.032N\right)\left(\frac{3}{5}\right)\\ F_{xon3}=-90.6192N\end{array}$

y component of net force on charge Q₃ :

  $\begin{array}{l}{F}_{yon3}=F_{1on3}\sin \theta +F_{2on3}\\ F_{yon3}=\left(-151.032N\right)\left(\frac{4}{5}\right)+\left(134.85N\right)\\ F_{yon3}=14.02N\end{array}$

Conclusion:

The x -component of the net electric force is $F_{xon3}=-90.6192N$ and y -component is $F_{yon3}=14.02N$.