Question

Two-point charges are arranged on the x-y coordinate system as follows; q₁= 3.0 nC at (0, 3m) and q₂= -9.0 nC at (0,0). Find the electric field at point on x-axis 4.0 m from the origin. 

Answer 1

The net electric field

$$\vec E=\vec E_1+\vec E_2$$

$$\vec E=k\frac{q_1}{r_1^3}\vec r_1+k\frac{q_2}{r_2^3}\vec r_2$$

$\vec r_1=(4,-3),\quad r_1=\sqrt{4^2+(-3)^2}=5$

$\vec r_2=(4,0),\quad r_2=\sqrt{4^2+0^2}=4$

$$\vec E=9\times 10^9\times \frac{3.0\times 10^{-9}}{5^3}(4,-3)\\ +9\times 10^9\times \frac{-9.0\times 10^{-9}}{4^3}(4,0)\\ =(0.864,-0.648)+(-5.063,0)\\ =(-4.2,-0.65)\:\rm N/C$$

The magnitud of field

$$E=\sqrt{E_x^2+E_y^2}=\sqrt{(-4.2)^2+(-0.65)^2}\\ =4.25\:\rm N/C$$