Question

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

Short answer

a) The electric flux through this cubical surface, if its edge length, $4\mathrm{cm}$is $3.62\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.

b) The electric flux through this cubical surface, if its edge length, $14\mathrm{cm}$is $51.1\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.

Step-by-step solution

The given data

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

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

Understanding the concept of the electric flux

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)

a) Calculation of the electric flux for edge length of 4 cm

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 $3.62\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.

b) Calculation of the electric flux for edge length of 14 cm

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 $51.1\hspace{0.33em}\mathrm{N}\cdot {\mathrm{m}}^2/\mathrm{C}$.