Question

A very long, thin wire has a uniform linear charge density of 99 µC/m. What is the electric field (in N/C) at a distance 3.0 cm from the wire? (Enter the magnitude.)

Answer 1

Answer:
`E=59.33* 10^6 N/C`

Step-by-step explanation:

Given

linear charge density
`(\lambda )=99 \mu C/m`

distance(r)=3 cm

Permittivity For free space
`\epsilon _0=8.85* 10^(-12) (C^2)/(N.m^2)`

Charge enclosed by gaussian surface

`q_(encl)=\lambda * l`

the flux through gaussian surface

`\phi =\oint EdA=E\left ( 2\pi rl\right )`

Also from Gauss law

`\phi =(q_(encl))/(\epsilon _0)`

Equate

`E\left ( 2\pi rl\right )=(q_(encl))/(\epsilon _0)`

`E=(\lambda )/(2\pi r\epsilon _0)`

`E=59.33* 10^6 N/C`