Question

A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very long line of charge with charge per unit length -6.00 μC/m is parallel to the x-axis at y = 10.
(a) What is the magnitude of the net electric field at point y2 = 0.200 m on the y-axis?

(b) What is the magnitude of the net electric field at point y3 = 0.616 m on the y-axis?

Answer 1

Answer:

a. E=6.5*10^{5}N/C

b. E=-4.7*10^{4}N/C

Explanation:

We can use the result of applying the Gauss theorem to a infinite line of charge

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

with r the perpendicular distance to the line of charge. The total field will be the sum of the contribution to the field by each line of charge.

E=E1+E2

we have

+8.00 μC/m

-6.00 μC/m

y=1m

(a) y2=0.2m

$E=\frac{8*10^{-6}\frac{C}{m}}{2\pi (0.2m)(8.85*10^{-12})\frac{C^2}{Nm^2}}+\frac{-6*10^{-6}\frac{C}{m}}{2\pi (0.8m)(8.85*10^{-12})\frac{C^2}{Nm^2}}\\\\E=7.9*10^{5}N/C-1.3*10^{5}N/C=6.5*10^{5}N/C$

(b) y3=0.616m

$E=\frac{8*10^{-6}\frac{C}{m}}{2\pi (0.616m)(8.85*10^{-12})\frac{C^2}{Nm^2}}+\frac{-6*10^{-6}\frac{C}{m}}{2\pi (0.384m)(8.85*10^{-12})\frac{C^2}{Nm^2}}\\\\E=2.33*10^{5}N/C-2.8*10^{5}N/C=-4.7*10^{4}N/C$

In this case the field is directed toward the line of charge placed in y=1m

Answer 2

Answer:

E =   = 5.8*10⁵N/C

E =-5.0*10⁵N/C

Explanation:

Answer:

Explanation:

Given;

λ₁ = 8.00uC/m  = 8.00 * 10⁻⁶

λ₂ = -6.00uC/m = -6.00*10⁻⁶

To calculate the magnitude of electric field at a point, Gauss's theorem of charge is used

It is given as

E = λ/(2πε₀r)

Where

   λ =is the charge density

   ε₀ =permissivity constant = 8.85*10⁻¹²

    r = distance

But E= E₁+ E₂

(a) At y = 0.2m, we have

E = 8* 10⁻⁶/(2*π *0.2*8.85*10⁻¹²) + -6* 10⁻⁶/(2* π * 0.8 * 8.85*10⁻¹²)

    =(8* 10⁻⁶/1.11* 10⁻¹¹) - (6* 10⁻⁶/(4.449 * 10⁻¹¹)

     =719251.12 - 134859.58

    = 583625.29

    = 5.8*10⁵N/C

(b) At y = 0.616, we have

E = 8* 10⁻⁶/(2*π *0.616*8.85*10⁻¹²) +

   -6* 10⁻⁶/(2* π * 0.38 *   8.85*10⁻¹²)

= (8* 10⁻⁶/3.4258 * 10⁻¹¹) -   (6* 10⁻⁶/2.1133*10⁻¹¹)

=            233522.10 - 283914.91

          = -50392.81

          =-5.0*10⁵N/C