91Ó°ÊÓ

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

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 Ï•.

Formulae:

The volume charge density of a material, $\mathrm{ÒÏ}=q/V$ (i)

The electric flux of a conducting Gaussian surface, $\mathrm{Ï•}=q/{\mathrm{Î\mu }}_0$ (ii)

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

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

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

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

Hence, the value of the flux is $3.62 \mathrm{N}â‹\dots {\mathrm{m}}^2/\mathrm{C}$.

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:

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

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

$$\begin{gathered} \mathrm{Ï•}=4.5×{10}^{-10}\mathrm{C}/\left(8.85×{10}^{-12}\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2\right) \\ 51.1 \mathrm{N}â‹\dots {\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the electric flux is $51.1 \mathrm{N}â‹\dots {\mathrm{m}}^2/\mathrm{C}$.