This question is incomplete, the complete question as well as the image of the problem designed are;
A finite line of charge with linear charge density λ = 3.35×10^−6 C/m and length L = 0.588 m is located along the x-axis (from x = 0 to x= L). A point charge of q = −6.22×10^−7 C is located al the point x0=1.42 m, y0=3.25 m.
Required: Find the electric field (magnitude and direction as measured from the +x-axis) at the point P which is located along the x-axis at xp=11.90 m.
The Coulomb force constant is k = 1/(4πE0) = 8.99×10^9 (N⋅m^2)/C^2
Answer: the electric field (magnitude and direction as measured from the +x ‑axis) is 88.2 N/C and 8.97
Explanation:
r = distance from the end of the rod = xp - L = 11.90 - 0.588 = 11.312 m
Q = charge on the rod = linear charge density x length = (3.35 x 10-6) (0.588) = 1.9698 x 10^-6
electric field by the rod is given as
Er = kQ/(r (xp)) = (8.99×10^9) (1.9698 x 10^-6) / (11.312 x 11.90)
= 17708.502 / 134.6128
= 131.55 N/C
BP = Xp - Xo = 11.90 - 1.42 = 10.48 m
AB = Yo = 3.25 m
Now using Pythagorean theorem
AP = √(AB^2 + BP^2) = √((10.48)^2 + (3.25)^2)
=√(109.8304 + 10.5625)
= √ ( 120.3929)
= 10.97
θ = tan^-1(AB/BP) = tan^-1( 3.25 / 10.48 )
= tan^-1 ( 0.3101 ) = 17.2
Electric field by the charge is given as
E' = kq/AP^2 = (8.99×10^9) (6.22×10^−7) / (10.97)^2
= 5591.78 / 120.3409
= 46.5 N/C
Net electric field along X-direction is given as
Ex = Er - E' Cos17.2
= 131.55 - (46.5 * 0.9553)
= 87.13 N/C
Net electric field along Y-direction is given as
Ey = E' Sin17.2
= 46.5 * 0.2957
= 13.75 N/C
N electric field is given as ;
E = √(Ex^2 + Ey^2)
= √((87.13)^2 + (13.75)^2)
= √ ( 7591.6369 + 189.0625 )
= √ ( 7780.6994 )
= 88.2 N/C
θ = angle = tan^-1(Ey/Ex)
= tan^-1 (13.75 / 87.13)
= tan^-1 ( 0.1578 )
= 8.97
Therefore, the electric field (magnitude and direction as measured from the +x ‑axis) is 88.2 N/C and 8.97
