Physics: An infinitely long nonconducting cylinder of radius R = 2.00 cm carries a uniform volume charge density of Calculate the electric field at distance r = 1.00 cm from the axis of the cylinder. (ε0 = 8.85 × 10-12 C2/N ∙ m2)

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An infinitely long nonconducting cylinder of radius

Ответ:

ramirezzairap2u4lh

18.01.2022

Physics

$E= 1.13*10^9\rho \: \: N/C$ where $\rho$ is the volume charge density.

Explanation:

Since the volume charge density is not given, I will just call it $\rho$ .

Gauss's law says

$\oint_S {E \cdot dA = \dfrac{Q_{inside}}{{\varepsilon _0 }}}.$

To evaluate this integral, we choose a cylindrical Gaussian surface which is concentric with the nonconducting cylinder.

Let be the radius of the Gaussian cylinder, and be its length, then the right side of the Gauss's law gives

$\oint_S E\cdot dA=E(2\pi rl)$

Notice that we only count the lateral area of the cylinder because the sides of the cylinder are perpendicular to , and therefore, give $E\cdot dA=0$ .

Now we turn to the right side of Gauss's law.

Since , the charge enclosed by the Gaussian surface is

$Q_{enc}= \rho*V= \rho*(\pi r^2)l$

which makes the right side

$\dfrac{Q_{enc}}{{\varepsilon _0 }}} = \dfrac{\rho \pi r^2l}{\varepsilon _0}$

Thus the Gauss's law becomes

$E(2\pi rl )=\dfrac{\rho \pi r^2l}{\varepsilon _0}$

simplifying a and solving for we get:

$\boxed{E= \dfrac{\rho r }{\varepsilon_0 }. }$

Putting in $\varepsilon_0 = 8.85*10^{-12} C^2/N*m^2$ and r=0.01m , we get:

$\boxed{ E= 1.13*10^9\rho \: \: N/C}$

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An infinitely long nonconducting cylinder of radius R = 2.00 cm carries a uniform volume charge density of Calculate the electric field at distance r = 1.00 cm from the axis of the cylinder. (ε0 = 8.85 × 10-12 C2/N ∙ m2)

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