Question

The net electric flux through each face of a die (singular of dice) has a magnitude in units of ${\mathbf{10}}^{\mathbf{3}}\mathbf{}{\mathbf{Nm}}^{\mathbf{2}}\mathbf{/}\mathbf{C}$that is exactly equal to the number of spots Non the face (1 through 6). The flux is inward for Nodd and outward for Neven. What is the net charge inside the die?

Short answer

The net charge inside the die is$2.66\times {10}^{-8}\mathrm{C}$ .

Step-by-step solution

Listing the given quantities

• The net electric flux =${10}^3\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$

Understanding the concept of Gauss law

Gauss law describes the relation between charge and electric field in a static situation. The equation for Gauss law is,

${\mathrm{\epsilon }}_0\mathrm{\Phi }={\mathrm{q}}_{\mathrm{enc}}$

Here, ${\mathit{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}}$is the net charge inside an imaginary closed surface and is the net flux of the electric field through the surface.

Net charge inside the die

Let ${\varnothing }_0={10}^{3}N.m^2/C$.

The net flux through the entire surface of the dice is given by:

$$\begin{gathered} \mathit{\varnothing }\mathit{}\mathit{=}\mathit{\sum }_{\mathit{n}\mathit{-}\mathit{1}}^{\mathit{6}}{\Phi }_{\mathit{n}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{\sum }_{\mathit{n}\mathit{-}\mathit{1}}^{\mathit{6}}{\left(-1\right)}^{\mathit{n}}n{\varphi }_{\mathit{0}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}{\varphi }_{\mathit{0}}\left(-1+2-3+4-5+6\right) \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{3}{\varphi }_{\mathit{0}} \end{gathered}$$

Thus, the net charge enclosed is:

role="math" localid="1657360360230" $$\begin{gathered} q={\epsilon }_0\varphi \\ =3{\epsilon }_0{\varphi }_0 \\ =3\left(8.85\times {10}^{-12}{C}^2/N.m^2\right)\left({10}^{3}N.m^2/C\right) \\ =2.66\times {10}^{-8}C \end{gathered}$$

Therefore, the net charge inside the die is$=2.66\times {10}^{-8}C$ .