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SOVA2
5 days ago
14

A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform li

ne of charge has charge per unit length λ2 = -2.26 μC/m and is parallel to the x-axis at y1 = 0.400 m . (Hint: You will need to know or derive using Gauss law the electric field by a uniform line charge)
Physics
1 answer:
inna [987]5 days ago
5 0

Answer:

a) E=228391.8 N/C

b) E=-59345.91 N/C

Explanation:

Gauss's law can be applied to determine the total electric field produced by both line charges.

\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}

Here, E1 and E2 represent the electric fields at distances r1 and r2 from each line charge, respectively.

The resultant electric field at position r can be expressed as:

E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})

a) for y=0.200m, with r1=0.200m and r2=0.200m:

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

b) for y=0.600m, with r1=0.600m and r2=0.200m:

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

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