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
[tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex]
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
[tex]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[/tex]
(b) y3=0.616m
[tex]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[/tex]
In this case the field is directed toward the line of charge placed in y=1m
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
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
