91影视

We are given charge = -3.00 nC is placed at the origin of an xy-coordinate system and a charge $q_2$= 2:00 nC is placed on the y-axis at y = 4,00 cm.

Required:

a) If a third charge, $q_3$=5.00 nC, is now placed at the point x=3.00 cm, y = 4.00 cm, as shown in the figure below, find the total force $F_{net}$exerted on this charge by the other two charges.

(b) Find the magnitude and direction of this force.

Solution:

It is easiest to calculate the electric force for each charge and that exerted on separately.

鈫扚or we could calculate its value by

$F_1=\frac{1}{4蟺{蔚}_0}\frac{\left|q_1{q}_3\right|}{r^2}$ (1)

Where r is the distance between and equals 5.0 cm. The term $\frac{1}{4蟺{蔚}_0}$equals $9.0脳{10}^9{\mathrm{Nmm}}^2/{\mathrm{C}}^2$. Now we can put our values for, and r into equation (1) to get

$F_1=\frac{1}{4蟺{蔚}_0}\frac{\left|q_1{q}_3\right|}{r^2}$

$\begin{array}{r}=\left(9.0脳{10}^9\mathrm{N}.{\mathrm{m}}^2/C^2\right)\left(\frac{\left|鈭�3.00脳{10}^{鈭�9}\mathrm{C}脳5.00脳{10}^{鈭�9}\right|}{(0.050\mathrm{m}{)}^2}\right)\\ =5.40脳{10}^{鈭�5}\mathrm{N}\end{array}$ 鈫扚or F鈧�

We can use equation (1) to get F鈧俠y putting our values for and r by 2.00 x C and 0.030 m respectively

$\begin{array}{r}{F}_2=\frac{1}{4蟺{蔚}_0}\frac{\left|q_2{q}_3\right|}{r^2}\\ =\left(9.0脳{10}^9{\mathrm{Nm}}^2/C^2\right)\left(\frac{\left|2.00脳{10}^{鈭�9}\mathrm{C}脳5.00脳{10}^{鈭�9}\right|}{(0.050\mathrm{m}{)}^2}\right)\\ =10脳{10}^{鈭�5}\mathrm{N}\end{array}$

Now we have two directions y and x as shown in the figure below for F鈧�, where

${\mathrm{F}}_1={\mathrm{F}}_{1\mathrm{x}}\mathrm{i}+{\mathrm{F}}_{1\mathrm{y}}\tilde{\mathrm{j}}$

While for F鈧� it is only in the x-direction

So let us combine all the electric fields in a direction and all the electric fields in y separately by

$\begin{array}{r}{F}_x=\left(鈭�F_{1x}+F_{2x}\right)i\\ F_y=\left(鈭�F_{1y}\right)j\end{array}$

Where F鈧俰s only in the x-direction as shown in the figure. Now we can get the combination of the electric forces that are exerted on by

$F_{\text{net}}=\sqrt{{\left(F_{1x}+F_{2x}\right)}^2+{\left(F_{1y}\right)}^2}$ (2)

From the figure below, we could get the value for each electric force in equation (2) by

role="math" localid="1668247314913" $\begin{array}{r}{F}_{1x}=鈭�F_1\cos 鈦�胃\\ F_{2x}=F_2\\ F_{1y}=鈭�F_1\sin 鈦�胃\end{array}$

where $\cos 鈦�胃=\frac{0.03}{0.05}$and $\sin 鈦�胃=\frac{0.04}{0.05}$. so it will be easy to calculate $F_{1x},F_{2x},F_{1y}$as next

$\begin{array}{c}{\mathrm{F}}_{1\mathrm{x}}=\left(鈭�5.40脳{10}^{鈭�5}\mathrm{N}脳\frac{0.03}{0.05}\right)=鈭�3.24脳{10}^{鈭�5}\mathrm{N}\\ {\mathrm{F}}_{2\mathrm{x}}=10.0脳{10}^{鈭�5}\mathrm{N}\\ {\mathrm{F}}_{1\mathrm{y}}=\left(5.40脳{10}^{鈭�5}\mathrm{N}脳\frac{0.04}{0.05}\right)=鈭�4.32脳{10}^{鈭�5}\mathrm{N}\end{array}$

now let us put these values for $F_{1x},F_{2x},F_{1y}$into equation (2) to get $F_{net}$

$\begin{array}{r}{\mathrm{F}}_{\mathrm{net}}=\sqrt{{\left({\mathrm{F}}_{1\mathrm{x}}+{\mathrm{F}}_{2\mathrm{x}}\right)}^2+{\left({\mathrm{F}}_{1\mathrm{y}}\right)}^2}\\ =8.02脳{10}^{鈭�5}\mathrm{N}\end{array}$

b) The magnitude of this force is as we calculate it.

$\left|\mathrm{F}\right|=8.02脳{10}^{鈭�5}\mathrm{N}$

To find the direction of the force, we could get the angle between the two forces and by

$\begin{array}{c}\tan 鈦�蠒=\frac{\left|F_y\right|}{\left|F_x\right|}\\ 蠒={\tan }^{鈭�1}鈦�\frac{\left|F_y\right|}{\left|F_x\right|}\\ 蠒={32.58}^{鈭�}\end{array}$

$\text{(Solve for}蠒\text{)}$

The direction of the force is with an angle $32.58掳$with +x direction.