Question

Consider a long cylindrical charge distribution of radius R = 17 cm with a uniform charge density of rho = 15 C/m3. Find the electric field at a distance r = 26 cm from the axis.

Answer 1

`E = 9.4*10^(10)N/C`

Step-by-step explanation:

We use Gauss's law which says

`$\int E\cdot dA = (Q_(enc))/(\varepsilon). $`
`(1)`

Now, for the cylindrical charge distribution the charge enclosed is

`Q_(enc) = \rho V`

where
`V` is the volume of the cylinder.

To evaluate Gauss's law, the Gaussian surface we choose is a cylinder concentric with the charged cylinder; therefore, equation
`(1)` becomes

`E (2\pi rL )=(\rho V)/(\varepsilon _o)`

`E (2\pi rL )=(\rho \pi R^2L)/(\varepsilon _o)`

`E =(\rho \pi R^2L)/( (2\pi rL )\varepsilon _o )`

`\boxed{E =(\rho R^2)/( 2\varepsilon _o r )}`

Putting in numerical values

`\rho = 15C/m^3`

`R = 17cm =0.17m`

`r = 26cm=0.26m`

`\varepsilon_0 =8.85*10^(-12)m^(-3)kg^(-1)s^4A^2}`

we get:

`E =(15 (0.17)^2)/( 2(8.85*10^(-12)) (0.26) )`

`\boxed{E = 9.4*10^(10)N/C}`