Question

A long cylinder (radius = 3.0 cm) is filled with a nonconducting material which carries a
uniform charge density of 1.3 uC/m². Determine the electric flux through a spherical
surface (radius = 2.0 cm) which has a point on the axis of the cylinder as its center.
a. 5.7 N·m2/C
b. 4.9 Nm2/C
c. 6.4 Nm2/C
d. 7.2 N·m2/C
e. 15 N·m2/C​

Answer 1

Answer:

The correct option is B

Explanation:

from the question we are told that

    The radius of the cylinder is $r = 3.0 cm = \frac{3}{100} = 0.03 \ m$

      The uniform charge density is  $\eta = 1.3 \mu C/m^2 = 1.3 *10^{-6} C/m^2$

      The radius of the spherical surface  $r_s = 2.0 \ cm = \frac{2}{100} = 0.02 \ m$

Generally the flux through the spherical surface  is mathematically represented as

        $\phi = \frac{Q}{\epsilon_o}$

Where  $\epsilon_o$ is the permittivity of free space with a value of  $\epsilon_o = 8.85*10^{-12} F/m$

Q is the quantity of charge which is mathematically represented as

        $Q = \eta * V$

Where V is the volume of the sphere which is mathematically represented as

      $V = \frac{4}{3} \pi r_s^3$

substituting values

      $V = \frac{4}{3}* 3.142 * 0.02^3$

      $V = 3.351*10^{-5} \ m^3$

So

    $Q = 1.3 *10^{-6} * 3.351*10^{-5}$

    $Q = 4.3569 *10^{-11} \ C$

So

     $\phi = \frac{ 4.3569 *10^{-11}}{8.85*10^{-12}}$

      $\phi = 4.92 N\cdot m ^2 /C$

 

Answer 2

The electric flux through the spherical surface is  $\rm 4.92\; N m^2 /C$

Gauss Law is used to determine the flux passing from a surface due to a point charge.

According to the given data

Radius of Cylinder = r = 3cm = 0.03 m

Let $\rho$ be the uniform charge density

$\rho = 1.3 \times 10^{-6}$ C/$m^2$

Radius of Spherical Surface = 0.02 m

The Electric flux  from spherical surface is given by equation (1)

$\rm \phi = \dfrac{Q}{\epsilon_o}....(1) \\Q = Net \; Charge \\\epsilon_o} = Permittivity\; of \; Free \; Space = 8.85 \times 10^{-12\; }F/m$

Also  $Q= \rho \times V....(2)$

$V = Volume \; of \; Sphere$

$V = 4/3 \times \pi \ r ^3$

where  r = Radius of the sphere

V = (4/3) $\times$ (3.14 )

Also from Equation (2) Q = 1.3 $\times$ $10^{-6}$ $\times 3.315\times 10^{-4}$

Q = 4.3569 $\times 10^{-11}$ C

So

$\phi = 4.3569 \times 10^{-11}/ 8.85 \times 10^{-12} \\\phi = 4.92 \; Nm^2 /C$

The electric flux through the spherical surface is $\rm 4.92\; N m^2 /C$

For more information refer to the link below

https://brainly.com/question/15168926