Question

A$\mathbf{6}\mathbf{.}\mathbf{20}\mathbf{\mu C}$ point charge is at the center of a cube with sides of length $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{m}$. (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides long $\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{m}$? Explain.

Short answer

(a) The electric flux through one of the six faces of the cube is ${\mathrm{F}}_{\mathrm{E}}=1.17\times {10}^5\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

(b) No change in electric flux.

Step-by-step solution

Definition of electric flux

The term electric flux may be defined as the application of electric field that may be thought of as the number of electric lines of force that intersect a given area.

Determine the electric flux

The electric flux can be calculated by the relation

${\mathrm{F}}_{\mathrm{E}}=\frac{{\mathrm{Q}}_{\mathrm{encl}}}{{\mathrm{\epsilon }}_0}$

Where$Q_{encl},{\epsilon }_0$ total charged enclosed and electric constant respectively.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{E}}=\frac{6.20\times {10}^{-6}\mathrm{C}}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2} \\ {\mathrm{F}}_{\mathrm{E}}=7.00\times {10}^5\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

And the electric flux from one side is calculated as

$$\begin{gathered} {\mathrm{F}}_{\mathrm{E}}=\frac{1}{6}\left(7.00\times {10}^5\right)\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \\ {\mathrm{F}}_{\mathrm{E}}=1.17\times {10}^5\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the electric flux through one of the six faces of the cube is ${\mathrm{F}}_{\mathrm{E}}=1.17\times {10}^5\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

Now the size of the cube change by the length 0.250m There is no any change in electric flux because it calculated by the charge enclosed in it.

Hence, no change in electric flux.