Answer:
[tex]E_{net} = (3.765\hat i - 1.207\hat j)kN/C[/tex]
Explanation:
Electric field due to long line charge on position of charge at x = 4 m is given as
[tex]E = \frac{2k\lambda}{r} \hat i[/tex]
so we have
[tex]\lambda = 0.30 \mu C/m[/tex]
now we have
[tex]E = \frac{2(9\times 10^9)(0.30 \mu C/m)}{4}[/tex]
[tex]E = 1350 N/C[/tex]
Now electric field due to the charge present at y = 2.0 m
[tex]E = \frac{kq}{r^2} \hat r[/tex]
[tex]E = \frac{(9\times 10^9)(6 \times 10^{-6})}{2^2 + 4^2}\times \frac{4\hat i - 2\hat j}{\sqrt{4^2 + 2^2}}[/tex]
[tex]E = 603.7 ( 4\hat i - 2\hat j)[/tex]
[tex]E = 2415\hat i - 1207.5 \hat j[/tex]
Now total electric field is given as
[tex]E_{net} = (1350\hat i) + (2415\hat i -1207.5\hat j)[/tex]
[tex]E_{net} = 3765\hat i - 1207.5 \hat j[/tex]
[tex]E_{net} = (3.765\hat i - 1.207\hat j)kN/C[/tex]
