Answer:[tex]E=75\ N/m[/tex]
Explanation:
Given
First charge of [tex]q_1=15\ nC[/tex] is placed at [tex]x=1.5\ m[/tex]
Second charge [tex]q_2=-20\ nC[/tex] is placed at [tex]y=-2\ m[/tex]
Electric field is given by
[tex]E=\frac{kq}{r^2}[/tex]
Electric field due to [tex]q_1[/tex] is away from it
[tex]E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}[/tex]
[tex]E_1=60\ N/m[/tex]
Electric field due to [tex]q_2[/tex]
[tex]E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}[/tex]
[tex]E_2=45\ N/m[/tex]
Net electric field will be vector addition of two
[tex]\vec{E_{net}}=\vec{E_1}+\vec{E_2}[/tex]
[tex]\vec{E_{net}}=-60\hat{i}-45\hat{j}[/tex]
Magnitude of Electric field is
[tex]E=\sqrt{60^2+45^2}[/tex]
[tex]E=75\ N/m[/tex]
