Question

A 8.08 uC point charge is at the center of a cube with sides of length 0.565 m.
Part A.) What is the electric flux through one of the six faces of the cube? (answer phi= ? N*m^2/C

Part B.) How would your answer to part A change if the sides were of length 0.150 ?

-could you explain your answer for Part B

Answer 1

Concepts and reason

The concept used here is Gauss’s law.

First use the gauss’s law to calculate the electric flux one face of the cube, then divide the flux through by 6 to get the flux through each of the six faces of the cube.

Finally, use the expression to calculate the flux through the cube when the length changes.

Fundamentals

The Gauss’s law states that charge stored inside is equal to the flux times the permittivity of free space. It is expressed as follows:

$q = {\varepsilon _0}\phi$

Here, q is the charge, ${\varepsilon _0}$ is the permittivity of free space, and $\phi$ is the flux through the surface.

(a)

Calculate the flux through one of the six faces of the cube.

The Gauss’s law states that charge stored inside is equal to the flux times the permittivity of free space. It is expressed as follows:

$q = {\varepsilon _0}\phi$

Here, q is the charge, ${\varepsilon _0}$ is the permittivity of free space, and $\phi$ is the flux through the surface.

Rewrite the Gauss’s law $q = {\varepsilon _0}\phi$ for flux $\phi$ .

$\phi = \frac{q}{{{\varepsilon _0}}}$

The expression to calculate the flux through the cube is,

$\phi = \frac{q}{{{\varepsilon _0}}}$

Here, q is the charge, ${\varepsilon _0}$ is the permittivity of free space, and $\phi$ is the flux through the surface.

The electric flux through one of the six faces of the cube is,

$\phi = \frac{q}{{6{\varepsilon _0}}}$

Substitute $8.08 \times {10^{ - 6}}\,{\rm{C}}$ for $q$ , $8.85 \times {10^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}$ for ${\varepsilon _0}$ in expression $\phi = \frac{q}{{6{\varepsilon _0}}}$ .

$\begin{array}{c}\\\phi = \frac{{8.08 \times {{10}^{ - 6}}\,{\rm{C}}}}{{6\left( {8.85 \times {{10}^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}}{{\rm{m}}^{\rm{2}}}} \right)}}\\\\ = 0.152 \times {10^6}\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}\\\end{array}$

(b)

The electric flux through one of the six faces of the cube is,

$\phi = \frac{q}{{6{\varepsilon _0}}}$

Here, q is the charge, ${\varepsilon _0}$ is the permittivity of free space, and $\phi$ is the flux through the surface.

From the expression, It is clear that the flux doesn’t depend on the side length.

Thus, the flux remains the same even the length of the cube sides changes.

Ans: Part a

The electric flux through one of the six faces of the cube is $152\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$ .

Part b

The electric flux through one of the six faces of the cube is $152\,{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/C}}$ .