Question

An infinite line charge of linear density λ = 0.30 µC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis at y = 2.0 m. The electric field at the point P on the x axis at x = 4.0 m is approximately.a. (4.2 kN/C) b. (4.2 kN/C) i + (0.64 kN/C) j c. (-0.96 kN/C) j d. (2.8 kN/C) i + (0.64 kN/C) j e. (5.2 kN/C) i = (2.3 kN/C) j

Answer 1

Answer:

$E_{net} = (3.765\hat i - 1.207\hat j)kN/C$

Explanation:

Electric field due to long line charge on position of charge at x = 4 m is given as

$E = \frac{2k\lambda}{r} \hat i$

so we have

$\lambda = 0.30 \mu C/m$

now we have

$E = \frac{2(9\times 10^9)(0.30 \mu C/m)}{4}$

$E = 1350 N/C$

Now electric field due to the charge present at y = 2.0 m

$E = \frac{kq}{r^2} \hat r$

$E = \frac{(9\times 10^9)(6 \times 10^{-6})}{2^2 + 4^2}\times \frac{4\hat i - 2\hat j}{\sqrt{4^2 + 2^2}}$

$E = 603.7 ( 4\hat i - 2\hat j)$

$E = 2415\hat i - 1207.5 \hat j$

Now total electric field is given as

$E_{net} = (1350\hat i) + (2415\hat i -1207.5\hat j)$

$E_{net} = 3765\hat i - 1207.5 \hat j$

$E_{net} = (3.765\hat i - 1.207\hat j)kN/C$