Question

Find the magnitude of the net electric field at the origin (created by charges q1 and q2).

Answer 1

The direction of the net magnetic field is $167.36^{\circ}$

How to find the magnitude?

$q_1=-4 \mathrm{nC} \text { at }(0.6,0.8)$

$q_2=6 \mathrm{nC}$ at (0.6,0)

$r_1=$Distance of $q_1$ from origin$=\sqrt{0.6^2+0.8^2}$

$r_2=$ Distance of $q_2$ from origin $=0.6$

$\mathrm{k}=$ Coulomb constant$=9$times $10^9 \mathrm{Nm}^2 / \mathrm{C}^2$

Electric field is given by

$&E_1=\frac{k q_1}{r_1^2}$

$&\Rightarrow E_1=\frac{9 \times 10^9 \times 4 \times 10^{-9}}{\sqrt{0.6^2+0.8^2}}$

$&\Rightarrow E_1=36 \mathrm{~N} / \mathrm{C}$

$&\theta=\tan ^{-1} \frac{0.8}{0.6}$

$&\Rightarrow \theta=53.13^{\circ}$

$&E_1=36 \cos 53.13^{\circ} \hat{i}+36 \sin 53.13^{\circ} \hat{j}$

$&\Rightarrow E_1=21.6 \hat{i}+28.8 \hat{j}$

$&E_2=\frac{k q_2}{r_2^2}$

$&\Rightarrow E_2=\frac{9 \times 10^9 \times 6 \times 10^{-9}}{0.6^2}$

$&\Rightarrow E_2=150 \mathrm{~N} / \mathrm{C}$

The charge $q_2$ is on the $\mathrm{x}$ axis itself and it is pointing towards the origin (left side) so the sign will be negative

$E_2=-150 \hat{i}$

Resultant electric field

$&E=E_1+E_2$

$&\Rightarrow E=21.6 \hat{i}+28.8 \hat{j}+(-150 \hat{i})$

$&\Rightarrow E=-128.4 \hat{i}+28.8 \hat{j}$

Magnitude of electric field is given by

$&|E|=\sqrt{(-128.4)^2+28.8^2}$

$&\Rightarrow|E|=121.6 \mathrm{~N} / \mathrm{C}$

Magnitude of the net electric field is $121.6 \mathrm{~N} / \mathrm{C}$

Direction is given by

$\theta=\tan ^{-1} \frac{128.4}{28.8}=77.36^{\circ}$

From the -x axis

$(90+77.36)^{\circ}=167.36^{\circ}$

The direction of the net magnetic field is $167.36^{\circ}$

To learn more about magnetic field, refer to:

brainly.com/question/26257705

#SPJ4