91Ó°ÊÓ

Electric flux,

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

$\mathrm{E}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{Ï\mu }}_{\mathrm{o}}\mathrm{A}}...1$

(a).

Assume $\mathrm{K}=\mathrm{R}$

For distance $\mathrm{r}<\mathrm{R}$(inside the sphere):

The charge enclosed is $+\mathrm{q}$.

So,

${\mathrm{Q}}_{\text{in}}=+\mathrm{q}$

The gaussian surface's area islocalid="1648761261922" $\mathrm{A}=4{\mathrm{Ï€°ù}}^2$.

Substitute all values in equation 1,

We get,

${\mathrm{E}}_{\mathrm{r}<\mathrm{R}}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{Ï\mu }}_{\mathrm{o}}\mathrm{A}}$

localid="1648761160531" $=\frac{\mathrm{q}}{{\mathrm{Ï\mu }}_{\mathrm{o}}\left(4{\mathrm{Ï€°ù}}^2\right)}$

$=\frac{1}{4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}}\frac{\mathrm{q}}{{\mathrm{r}}^2}$

Because the electric field points outward from the positive charge, the electric field's direction is outward from the sphere.

(b).

Assume $\mathrm{K}=\mathrm{R}$

For distance $\mathrm{r}>\mathrm{R}$(outside the sphere):

The charge enclosed is $+\mathrm{q}$and $-2\mathrm{q}$.

So,

${\mathrm{Q}}_{\text{in}}=+\mathrm{q}+(-2\mathrm{q})=-\mathrm{q}$

The gaussian surface's area is role="math" localid="1648761665087" $\mathrm{A}=4{\mathrm{Ï€°ù}}^2$.

Substitute all values in equation$1$,

We get,

${\mathrm{E}}_{\mathrm{r}>\mathrm{R}}=\frac{{\mathrm{Q}}_{\text{in}}}{{\mathrm{Ï\mu }}_{\mathrm{o}}\mathrm{A}}$

$=\frac{-\mathrm{q}}{{\mathrm{Ï\mu }}_{\mathrm{o}}\left(4{\mathrm{Ï€°ù}}^2\right)}$

role="math" localid="1648761787492" $=\frac{1}{4{\mathrm{Ï€Ï\mu }}_{\mathrm{o}}}\frac{-\mathrm{q}}{{\mathrm{r}}^2}$

The electric field is directed inward the sphere because it is directed inward the negative charge.