Question

A finite line of charge with linear charge density ????=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 at the point x0=1.42 m, y0=3.25 m.
Required:
Find the electric field (magnitude and direction as measured from the +x ‑axis) which is located along the x ‑axis at x=11.90 m. The Coulomb force constant is k=1/(4π/????0)=8.99×10^9 (N⋅m^2)/C^2.

Answer 1

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