A uniform charge density of $\mathbf{500}\mathbf{\hspace{0.33em}}\mathbf{nC}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$is distributed throughout a spherical volume of radius$\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$. Consider a cubical Gaussian surface with its center at the center of the sphere. What is the electric flux through this cubical surface if its edge length is

(a)$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$and

(b)$\mathbf{14}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$?

a) Uniform charge density of$\mathrm{\rho }=500\mathrm{nC}/{\mathrm{m}}^3$

b) Radius spherical volume,$\mathrm{R}=6.00\mathrm{cm}$

Using the concept of Volume charge density, the value of the net charge q is calculated for both cases. Then using the flux concept from the Gauss theorem, we can get the required value of the electric flux ϕ.

Formulae:

The volume charge density of a material, $\rho =q/V$ (i)

The electric flux of a conducting Gaussian surface, $\varphi =q/{\epsilon }_0$ (ii)

The cube is totally within the spherical volume, so the charge enclosed is given using equation (i) as:

$$\begin{gathered} {\mathrm{q}}_{\mathrm{enc}}={\mathrm{pV}}_{\mathrm{cube}} \\ =\left(500\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^3\right){\left(0.040\mathrm{m}\right)}^3 \\ =3.2\times {10}^{-11}\mathrm{C} \end{gathered}$$

Now, using the above charge value, the electric flux through the surface is given using equation (ii) as:

$$\begin{gathered} \mathrm{\varphi }=3.2\times {10}^{-11}\mathrm{C}/\left(8.85\times {10}^{-12}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right) \\ =3.62\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the flux is $3.62\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.

Now the sphere is totally contained within the cube (note that the radius of the sphere is less than half the side-length of the cube). Thus, the total charge is:

$$\begin{gathered} {\mathrm{q}}_{\mathrm{enc}}={\mathrm{pV}}_{sphere} \\ =\frac{4\pi }{3}\left(500\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^3\right){\left(0.06\mathrm{m}\right)}^3 \\ =4.5\times {10}^{-10}\mathrm{C} \end{gathered}$$

Now, using the above charge value, the electric flux through the surface is given using equation (ii) as:

$$\begin{gathered} \mathrm{\varphi }=4.5\times {10}^{-10}\mathrm{C}/\left(8.85\times {10}^{-12}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right) \\ 51.1\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the electric flux is $51.1\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.