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

`E_(net) = (3.765\hat i - 1.207\hat j)kN/C`

Step-by-step explanation:

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

`E = (2k\lambda)/(r) \hat i`

so we have

`\lambda = 0.30 \mu C/m`

now we have

`E = (2(9* 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 = (kq)/(r^2) \hat r`

`E = ((9* 10^9)(6 * 10^(-6)))/(2^2 + 4^2)* (4\hat i - 2\hat j)/(√(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`