91影视

The electric field at charge is

$E=\frac{k\left|q\right|}{r^2}$

In the diagram below, the parts of the magnetic charge and the particles are depicted:

Equating the angle from the diagram as

$\begin{array}{l}\sin 鈦�胃=\frac{3}{\sqrt{34}}\\ \sin 鈦�胃=0.51.\\ \cos 鈦�胃=\frac{5}{\sqrt{34}}\\ \cos 鈦�胃=0.84.\end{array}$

The electric field component as

$E_x=E_3鈭�E_2\cos 鈦�胃$

Equating as

$E_x=\frac{k\left|q_3\right|}{r_3^2}鈭�\frac{k\left|q_2\right|}{r_2^2}\cos 鈦�胃$

Solving as,

$\begin{array}{l}{E}_x=\frac{\left(9脳{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\right)\left(5.0\mathrm{nC}\left(\frac{{10}^{鈭�9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(5\mathrm{cm}\left(\frac{{10}^{鈭�2}\mathrm{m}}{1\mathrm{cm}}\right)\right)}^2}鈭�\frac{\left(9脳{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\right)\left(10.0\mathrm{nC}\left(\frac{{10}^{鈭�9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{\left(0.51\right)}\\ E_x=1.8脳{10}^4\mathrm{N}/\mathrm{C}鈭�2.28脳{10}^4\mathrm{N}/\mathrm{C}\\ E_x=鈭�0.48脳{10}^4\mathrm{N}/\mathrm{C}\\ E_x{\left.=鈭�4.8脳{10}^3\mathrm{N}/\mathrm{C}\left(\frac{{10}^{鈭�2}\mathrm{m}}{1\mathrm{cm}}\right)\right)}^2\end{array}$

The field strength as

$E_y=E_1鈭�E_2\sin 鈦�胃$

The equation as

$E_y=\frac{k\left|q_1\right|}{r_1^2}鈭�\frac{k\left|q_2\right|}{r_2^2}\sin 鈦�胃$

Solving as

$\begin{array}{l}{E}_y=\frac{\left(9脳{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\right)\left(10\mathrm{nC}\left(\frac{{10}^{鈭�9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(3\mathrm{cm}\left(\frac{{10}^{鈭�2}\mathrm{m}}{1\mathrm{cm}}\right)\right)}^2}鈭�\frac{\left(9脳{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\right)\left(10.0\mathrm{nC}\left(\frac{{10}^{鈭�9}\mathrm{C}}{1\mathrm{nC}}\right)\right)}{{\left(\sqrt{34}\mathrm{cm}\left(\frac{{10}^{鈭�2}\mathrm{m}}{1\mathrm{cm}}\right)\right)}^2}\left(0.84\right)\\ E_y=10脳{10}^4\mathrm{N}/\mathrm{C}鈭�1.35脳{10}^4\mathrm{N}/\mathrm{C}\\ E_y=8.65脳{10}^4\mathrm{N}/\mathrm{C}.\end{array}$

The magnitude at field is

$\left|E\right|=\sqrt{E_x^2+E_y^2}$

Solving as

$\begin{array}{l}\left|E\right|=\sqrt{{\left(鈭�4.8脳{10}^3\mathrm{N}/\mathrm{C}\right)}^2+{\left(8.65脳{10}^4\mathrm{N}/\mathrm{C}\right)}^2}\\ \left|E\right|=8.6脳{10}^4\mathrm{N}/\mathrm{C}.\end{array}$

The angle as

$\tan 鈦�胃=\frac{E_y}{E_x}$

$胃={\tan }^{鈭�1}鈦�\left(\frac{E_y}{E_x}\right)$

Solving as

$\begin{array}{l}胃={\tan }^{鈭�1}鈦�\left(\frac{8.6脳{10}^4\mathrm{N}/\mathrm{C}}{鈭�4.8脳{10}^3\mathrm{N}/\mathrm{C}}\right)\\ 胃=鈭�{86.81}^{鈭�}.\end{array}$

The angle in $x-$axis as positive and as clockwise.

$\begin{array}{l}胃={180}^{鈭�}鈭�{87}^{鈭�}\\ 胃={93}^{鈭�}.\end{array}$