91影视

Given $Q=鈭�1.75渭\mathrm{C},m=5.0\mathrm{g},a=324{\mathrm{ms}}^{鈭�2}\text{and}L=3.0\mathrm{cm}$

The net force is in the direction of acceleration so we must arrange the system as shown in the figure to make the net force in direction of +y.

The net force according to Newton鈥檚 second law of motion

$\begin{array}{r}{F}_{net}=ma\\ F_{ntt}=5脳{10}^{鈭�3}脳324\\ =1.62N鈥�.=>\left(1\right)\end{array}$

Also, the net force is

$F_{\text{net}}=F_1\sin 鈦�\left(胃\right)+F_2\cos 鈦�\left(胃\right)鈥�鈥�鈬�\left(2\right)$

From (1) and (2)

$F_1\sin 鈦�\left(胃\right)+F_2\cos 鈦�\left(胃\right)=1.62鈥�鈥�\left(3\right)$

Net Force in the x-direction is

$F_1\cos 鈦�\left(胃\right)=F_2\sin 鈦�\left(胃\right)鈥�鈥�=>\left(4\right)$

From (3) and (4)

$F_2\tan 鈦�\left(胃\right)\sin 鈦�\left(胃\right)+F_2\cos 鈦�\left(胃\right)=1.62$

$F_2=\frac{1.62}{\tan 鈦�\left(胃\right)\sin 鈦�\left(胃\right)+\cos 鈦�\left(胃\right)}鈥�.鈬�\left(5\right)$

The angle is

$胃=\arcsin 鈦�\left(\frac{2.25}{3}\right)={48.6}^{鈭�}$

Substitutes in (5) yields

$F_2=1.07\mathrm{N}$

Substitutes in (4) yields

$\begin{array}{r}{F}_1=\frac{1.07\cos 鈦�\left(48.6\right)}{\sin 鈦�\left(48.6\right)}\\ =0.94N\end{array}$

Hence the $F_2=1.07\mathrm{N}\text{and}{F}_1=0.94\mathrm{N}$

The forces are electric forces between charges so to get q1

$F_1=\frac{kQ{q}_1}{L^2鈥�..}鈬�$(6)

Substitutes in (6) yields

$\begin{array}{r}{q}_1=\frac{0.94脳(0.03{)}^2}{9脳{10}^9脳1.75脳{10}^{鈭�6}}\\ =5.37脳{10}^{鈭�8}\mathrm{C}\end{array}$

And hence the charge is positive.

$F_2=\frac{kQ{q}_2}{L^2}鈥�..>\left(7\right)$

Substitute in (7) yields

$\begin{array}{r}{q}_2=\frac{1.07脳(0.03{)}^2}{9脳{10}^9脳1.75脳{10}^{鈭�6}}\\ =6.11脳{10}^{鈭�8}\mathrm{C}\end{array}$

And hence the charge is negative so $q_3=6.11脳{10}^{鈭�8}\mathrm{C}$.