Answer:
[tex]121.6\ \text{N/C}[/tex]
[tex]167.36^{\circ}[/tex]
Explanation:
[tex]q_1=-4\ \text{nC}\ \text{at }(0.6,0.8)[/tex]
[tex]q_2=6\ \text{nC}\ \text{at }(0.6,0)[/tex]
[tex]r_1[/tex] = Distance of [tex]q_1[/tex] from origin = [tex]\sqrt{0.6^2+0.8^2}[/tex]
[tex]r_2[/tex] = Distance of [tex]q_2[/tex] from origin = [tex]0.6[/tex]
k = Coulomb constant = [tex]9\times 10^9\ \text{Nm}^2/\text{C}^2[/tex]
Electric field is given by
[tex]E_1=\dfrac{kq_1}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 4\times 10^{-9}}{\sqrt{0.6^2+0.8^2}}\\\Rightarrow E_1=36\ \text{N/C}[/tex]
[tex]\theta=\tan^{-1}\dfrac{0.8}{0.6}\\\Rightarrow\theta=53.13^{\circ}[/tex]
[tex]E_1=36\cos53.13^{\circ}\hat{i}+36\sin53.13^{\circ}\hat{j}\\\Rightarrow E_1=21.6\hat{i}+28.8\hat{j}[/tex]
[tex]E_2=\dfrac{kq_2}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 6\times 10^{-9}}{0.6^2}\\\Rightarrow E_2=150\ \text{N/C}[/tex]
The charge [tex]q_2[/tex] is on the x axis itself and it is pointing towards the origin (left side) so the sign will be negative
[tex]E_2=-150\hat{i}[/tex]
Resultant electric field
[tex]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}[/tex]
Magnitude of electric field is given by
[tex]|E|=\sqrt{(-128.4)^2+28.8^2}\\\Rightarrow |E|=121.6\ \text{N/C}[/tex]
Magnitude of the net electric field is [tex]121.6\ \text{N/C}[/tex]
Direction is given by
[tex]\theta=\tan^{-1}\dfrac{128.4}{28.8}=77.36^{\circ}[/tex]
From the -x axis
[tex](90+77.36)^{\circ}=167.36^{\circ}[/tex]
The direction of the net magnetic field is [tex]167.36^{\circ}[/tex]
