91Ó°ÊÓ

(a).

The amount of electrical field that travels through a closed surface is stated because the electric flux.

The electric field through a surface is expounded to the charge inside the surface, in line with Gauss's law.

Electric flux is,

${\mathrm{Φ}}_{\mathrm{e}}=\mathrm{EA}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{Ï\mu }}_{\mathrm{o}}}$

${\mathrm{Q}}_{\text{in}}={\mathrm{Ï\mu }}_{\mathrm{o}}\mathrm{EA}$

localid="1648745447364" ${\mathrm{Q}}_{\text{in}}=4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}{\mathrm{Er}}^2.....1$

The gaussian surface's radius is localid="1648747531003" $8\mathrm{cm}$.

Because the electrical field is contained within the hollow sphere, it's negative.

$\mathrm{E}=-15000\mathrm{N}/\mathrm{C}$.

Substitute values in equation localid="1648747540022" $1$,

We get,

${\mathrm{Q}}_{\text{in}}=4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}{\mathrm{Er}}^2$

$=4\mathrm{π}\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)(-15000\mathrm{N}/\mathrm{C})(0.08\mathrm{m}{)}^2$

$=-10.7×{10}^{-9}\mathrm{C}=-10.7\mathrm{nC}$

(b).

The electric charge on the inner sphere's external surface generates a electric charge of the identical magnitude on the hollow sphere's interior surface.

The charge on the hollow sphere's inner surface is the image of the charge inside the small sphere, except it points within the other direction.

So,

${\mathrm{Q}}_{\text{in}}=-(-10.7\mathrm{nC})=+10.7\mathrm{nC}$

(c).

The radius of the gaussian surface is $\mathrm{r}=17\mathrm{cm}$

The electric field emanating from the hollow spherical is positive.

$\mathrm{E}=15000\mathrm{N}/\mathrm{C}$

Substitute values in equation localid="1648747517181" $1$,

We get,

${\mathrm{Q}}_{\mathrm{in}}=4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}{\mathrm{Er}}^2$

$=4\mathrm{π}\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)(15000\mathrm{N}/\mathrm{C})(0.17\mathrm{m}{)}^2$

$=48.2×{10}^{-9}\mathrm{C}=48.2\mathrm{nC}$