Respuesta :
Answer:
a) E=228391.8 N/C
b) E=-59345.91N/C
Explanation:
You can use Gauss law to find the net electric field produced by both line of charges.
[tex]\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}[/tex]
Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.
The net electric field at point r will be:
[tex]E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})[/tex]
a) for y=0.200m, r1=0.200m and r2=0.200m:
[tex]E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C[/tex]
b) for y=0.600m, r1=0.600m, r2=0.200m:
[tex]E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C[/tex]
a) The electric field strength for case 1 will be 228391.8 N/C
b) The electric field strength for case 2 will be 59345.91 N/C
What is gauss law?
The total electric flux out of a closed surface is equal to the charge contained divided by the permittivity,
According to Gauss Law. The electric flux in a given area is calculated by multiplying the electric field by the area of the surface projected in a plane perpendicular to the field.
The given data in the problem is;
λ₁ is the charge per unit length for charge 1 = 4.80 μC/m
λ₂ is the charge per unit length for charge 2 =-2.26 μC/m
y₁ is the distance from charge 1 =0.400 m.
From the gauss law, the net electric field is produced by both lines of charges will be .;
[tex]\rm E= \frac{\lambda_}{2 \pi \epsilon_0 r_2} \\\\ \rm E_1= \frac{\lambda_1}{2 \pi \epsilon_0 r_1} \\\\ \rm E_2= \frac{\lambda_2}{2 \pi \epsilon_0 r_2}[/tex]
The total electric field will be the sum of the electric field for the charge 1 and2;
[tex]\rm E=E_1+E_2 \\\\ \rm E=\frac{\lambda_1}{2 \pi \epsilon_0 r_1}+\frac{\lambda_2}{2 \pi \epsilon_0 r_2} \\\\[/tex]
The net electric field for y=0.200m, r₁=0.200m, and r₂=0.200 m will be;
[tex]\rm E=\frac{4.80 \times 10^{-6}}{2 \times 3.14 8.85 \times 10^{-12} \times 0.200}-\frac{2.26 \times 10^{-6}}{2 \times 3.14 \times 8.85 \times 10^{-12}} \\\\\rm E=228391.8 \ N/C[/tex]
The net electric field for y=0.600m, r₁=0.600m, r₂=0.200 m will be;
[tex]\rm E=\frac{4.80 \times 10^{-6}}{2 \times 3.14 8.85 \times 10^{-12} \times 0.600}-\frac{2.26 \times 10^{-6}}{2 \times 3.14 \times 8.85 \times 10^{-12}} \\\\\rm E=-59345.51 \ N/C[/tex]
Hence the electric field strength for case 1 and case 2 will be 228391.8 N/C and 59345.91 N/C respectively.
To learn more about the gauss law refer to the link;
https://brainly.com/question/2854215
