A particle with charge 5.0 ° C is placed at the corner of a cube.The total electric flux through all sides of the cube is: A)0 N.m²/C B)7.1 * 10⁴ N.m²/C C)9.4 * 10⁴ N.m²/C D)1.4 * 10⁵ N.m²/C E)5.6 * 10⁵ N.m²/C

Question

Quizplus · Accepted Answer

The electric flux through a closed surface is given by the formula:

$\phi_E = \oint \vec{E}\cdot d\vec{A}$

where $\vec{E}$ is the electric field and $d\vec{A}$ is an infinitesimal area element.

For a cube, there are 6 faces, each with area $a^2$, where $a$ is the length of a side of the cube. The electric field due to the charged particle at one corner of the cube is radial, and it points outward. The magnitude of the electric field due to a point charge $q$ at a distance $r$ from the charge is given by Coulomb's law:

$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

where $\epsilon_0$ is the permittivity of free space.

To find the electric flux through one face of the cube, we need to find the component of the electric field that is perpendicular to the face. Since the electric field is radial and points outward, the component of the electric field that is perpendicular to one face is the same as the magnitude of the electric field at the center of the face. The distance from the corner of the cube to the center of a face is $a/\sqrt{2}$. Therefore, the electric field at the center of a face is:

$E_{face} = \frac{1}{4\pi\epsilon_0}\frac{q}{(a/\sqrt{2})^2} = \frac{1}{4\pi\epsilon_0}\frac{q}{a^2/2}$

The electric flux through one face is then:

$\phi_{E,face} = E_{face} A_{face} = \frac{1}{4\pi\epsilon_0}\frac{q}{a^2/2} a^2 = \frac{q}{2\epsilon_0}$

Since there are 6 faces, the total electric flux through all sides of the cube is:

$\phi_E = 6\phi_{E,face} = 6\frac{q}{2\epsilon_0} = \frac{3q}{\epsilon_0}$

Plugging in the given charge value of $5.0\mu C$, we get:

$\phi_E = \frac{3(5.0	imes 10^{-6} C)}{8.85	imes 10^{-12} C^2/N\cdot m^2} \approx 7.1	imes 10^{10} N\cdot m^2/C$

Therefore, the answer is (B).

A Particle with Charge 5