Answer:
23.1 N/C
Explanation:
OP = 3 m , OQ = 4 m
[tex]PQ = \sqrt{4^{2}+3^{2}}=5 m[/tex]
q = - 8 nC, Q = 75 nC
Electric field at P due to the charge Q is
[tex]E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C[/tex]
Electric field at P due to the charge q is
[tex]E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C[/tex]
According to the diagram, tanθ = 3/4
Resolve the components of E1 along x axis and along y axis.
So, Electric field along X axis, Ex = - E1 Cos θ
Ex = - 27 x 4 / 5 = - 21.6 N/C
Electric field along y axis, Ey = E1 Sinθ - E2
Ey = 27 x 3 /5 - 8 = 8.2 N/C
The resultant electric field at P is given by
[tex]E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C[/tex]
