Answer :
Answer:
Explanation:
q1 = 10 nC at x = 4 cm
q2 = 5 nC at y = - 2 cm
Let E1 be the electric field due to q1 at origin
[tex]E_{1}=\frac{Kq_{1}}{x^{2}}[/tex]
[tex]E_{1}=\frac{9\times 10^{9}\times 10\times 10^{-9}}{0.04^{2}}[/tex]
E1 = 56250 N/C
Let E2 be the electric field due to q2 at origin
[tex]E_{2}=\frac{Kq_{2}}{y^{2}}[/tex]
[tex]E_{2}=\frac{9\times 10^{9}\times 5\times 10^{-9}}{0.02^{2}}[/tex]
E2 = 112500 N/C
The two electric fields are perpendicular to each other so the net electric field is
[tex]E = \sqrt{E_{1}^{2}+E_{2}^{2}}[/tex]
[tex]E = \sqrt{56250^{2}+112500^{2}}[/tex]
E = 1.26 x 10^6 N/C
Let the angle is Ф.
tan Ф = E2 / E1
tan Ф = 2
Ф = 63.4 ° from negative X axis upwards .