Answer to Question #188159 in Electricity and Magnetism for Rabeet Ul Hassan

Question #188159

Consider the electric field ๐ธ = ๐‘Ž๐‘Ÿ

๐‘Ÿ๐œŒ๐‘™

4๐œ‹๐œ–0

โˆซ

๐‘‘๐‘ง

โ€ฒ

(๐‘Ÿ

2+๐‘งโ€ฒ

2)

3/2

+โˆž

โˆ’โˆž

due to the uniform line charge density ฯl

= 12 ยตC/m of an infinitely long straight conducting wire at a distance r. Find out the numeric value ofย 

r. Remember that prime variables corresponds to the source quantity.


1
Expert's answer
2021-05-03T10:28:35-0400

Given,

Electric field "E=ar"

"E=\\frac{r\\rho l}{4\\pi \\epsilon_o}\\int_{-\\infty}^{\\infty} \\frac{dz}{(r^2+z^2)^{3\/2}}"


"=\\frac{ar\\rho l}{4\\pi \\epsilon_o}\\times [\\frac{z}{r^2(z^2+r^2)}]_{-\\infty}^{\\infty}"


"=\\frac{ar\\rho l}{4\\pi \\epsilon_o}\\times (\\frac{2}{r^2})"

So,

"ar=\\frac{r\\rho l}{4\\pi \\epsilon_o}\\times (\\frac{2}{r^2})"


"\\Rightarrow" "r=\\sqrt{\\frac{2\\rho l}{4\\pi \\epsilon_o a}}"


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