91Ó°ÊÓ

Consider the formula for the gauss law for the electric displacement as follows:

$âˆ\circledR \stackrel{\mathbf{→}}{\mathbf{D}}·\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{a}}={\mathbf{Q}}_{\mathbf{encl}}$

Here, D is the electric displacement, is the area of element and is the charge that is enclosed.

Consider the formula for the charge in terms of the volume charge density is as follows:

$\mathbf{Q}\mathbf{=}\left(\frac{\mathbf{4}{\mathbf{Ï€°ù}}^{\mathbf{3}}}{\mathbf{3}}\right)\mathbf{ÒÏ}$

Write the expression for the electric potential in terms of the electric field as;

$\mathbf{V}\mathbf{=}\mathbf{-}∫\stackrel{\mathbf{→}}{\mathbf{E}}·\mathbf{d}\stackrel{\mathbf{→}}{\mathbf{l}}$

Consider the expression for the electric field as:

$\stackrel{→}{E}=\frac{\stackrel{→}{D}}{\mathrm{Î\mu }}$

Here, is the dielectric constant.

Consider the electric potential expression as:

$V=-∫\stackrel{→}{E}·dl$

Solve for the charge inside the sphere as:

$Q=\left(\frac{4\mathrm{Ï€}{r}^3}{3}\right)\mathrm{ÒÏ}$

Consider for , rewrite the equation as:

$Q=\left(\frac{4\mathrm{Ï€}{R}^3}{3}\right)\mathrm{ÒÏ}$

Consider the electric displacement by the Gauss law is:

$$\begin{gathered} âˆ\circledR \stackrel{→}{D}·d\stackrel{→}{a}=Q_{encl} \\ D·A=Q_{end} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} D\left(4\mathrm{Ï€}{r}^2\right)=\left(\frac{4\mathrm{Ï€}{r}^3}{3}\right)\mathrm{ÒÏ} \\ \stackrel{→}{D}=\frac{\mathrm{ÒÏ}r}{3}\hat{r}\backslash \end{gathered}$$

Consider the expression for the electric field is given as:

$\stackrel{→}{E}=\frac{\stackrel{→}{D}}{\mathrm{Î\mu }}$

Substitute and rewrite as:

$\stackrel{→}{E}=\frac{\mathrm{ÒÏ}r}{3\mathrm{Î\mu }}\hat{r}$

Consider the electric displacement is needed to determine the field outside the sphere by the gauss law.

Solve for the electric displacement as:$\frac{\mathrm{ÒÏ}{R}^2}{3{\mathrm{Î\mu }}_0}\left(1+\frac{1}{2{\mathrm{Î\mu }}_r}\right)$

$$\begin{gathered} D\left(4\mathrm{Ï€}{r}^2\right)=\left(\frac{4\mathrm{Ï€}{R}^3}{3}\right)\mathrm{ÒÏ} \\ \stackrel{→}{D}=\frac{\mathrm{ÒÏ}{R}^3}{3r^2}\hat{r} \end{gathered}$$

Then, write the expression for the electric field as:

$\stackrel{→}{E}=\frac{\mathrm{ÒÏ}{R}^3}{3{\mathrm{Î\mu }}_0{r}^2}\hat{r}$

Determine the electric potential inside and outside sphere as follows:

$$\begin{gathered} V={∫}_{∞}^R\frac{\mathrm{ÒÏ}{R}^3}{3{\mathrm{Î\mu }}_0{r}^2}dr-{∫}_R^0\frac{\mathrm{ÒÏ}r}{3\mathrm{Î\mu }}dr \\ =-\frac{\mathrm{ÒÏ}}{3{\mathrm{Î\mu }}_0}\left(R^3{\left[-\frac{1}{r^2}\right]}_{∞}^R+\frac{1}{2{\mathrm{Î\mu }}_r}{\left[r^2\right]}_R^0\right) \\ =\frac{\mathrm{ÒÏ}{R}^2}{3{\mathrm{Î\mu }}_0}\left(1+\frac{1}{2{\mathrm{Î\mu }}_r}\right) \end{gathered}$$

Therefore, the electric potential at the centre off the sphere is .