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

Answer:

$E = 9.4*10^{10}N/C$

Explanation:

We use Gauss's law which says

$\int E\cdot dA = \frac{Q_{enc}}{\varepsilon}.$     $(1)$

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

$E (2\pi rL )=\dfrac{\rho V}{\varepsilon _o}$

$E (2\pi rL )=\dfrac{\rho \pi R^2L}{\varepsilon _o}$

$E =\dfrac{\rho \pi R^2L}{ (2\pi rL )\varepsilon _o }$

$\boxed{E =\dfrac{\rho R^2}{ 2\varepsilon _o r }}$

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:

$E =\dfrac{15 (0.17)^2}{ 2(8.85*10^{-12}) (0.26) }$

$\boxed{E = 9.4*10^{10}N/C}$