91Ó°ÊÓ

The distance, $r>r_2$ , at $V=0$

Here, the electric potential are equal.

The net charge on the thick spherical shellis Q.

Uniform volume charge densityis P,

At infinity the electric potential,$\mathit{V}\mathbf{=}\mathbf{0}$

Gauss's law for electricity states that the electric flux$\mathit{Φ}$across any closed surface is proportional to the net electric charge q enclosed by the surface; that is,

$\mathit{Φ}\mathbf{=}\raisebox{1ex}{$\mathbf{\text{q}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{ε}}}_{\mathbf{\text{0}}}$}\right.$

Here, ${\mathbf{\text{ε}}}_{\mathbf{\text{0}}}$is the electric permittivity of free space and has a value of $\mathbf{8}\mathbf{.85}\mathbf{×}{\mathbf{10}}^{\mathbf{−}\mathbf{12}}{\mathbf{\text{C}}}^{\mathbf{2}}\mathbf{/}\mathbf{\text{N}}\mathbf{â‹\dots }{\mathbf{\text{m}}}^{\mathbf{2}}$.

Formulae:

The electric potential is define as below.

$V\left(r\right)=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}â‹\dots \frac{Q}{r}$

The electric flux is define as,

$f=\underset{}{\overset{}{}}Eâ‹\dots dA=\frac{1}{{\mathrm{Î\mu }}_o}q$

Where, V is potential energy, R is distance between the point charges, q is charge, E is electric potential, and A is the area.

If $r>r_2$:
The electric potential is given as:

$V\left(r\right)=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}â‹\dots \frac{Q}{r}$

Hence, the electric potential at $r>r_2$ is $V\left(r\right)=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}â‹\dots \frac{Q}{r}$.

If$r_2>r>r_1$,
Since the Gaussian surface lies inside the shell,
The volume of the shell is,

$V=\frac{4}{3}\mathrm{π}(r_2^3−r_1^3)$
So, the volume charge density is,

$\mathrm{ÒÏ}=\frac{Q}{V}$

The charge enclosed by the shell is,

role="math" localid="1662605913199" $\begin{array}{c}q=\mathrm{ÒÏ}V\\ =\frac{3Q}{4\mathrm{Ï€}(r_2^3−r_1^3)}×\frac{4}{3}\mathrm{Ï€}(r^3−r_1^3)\\ =Q\left(\frac{r^3−r_1^3}{r_2^3−r_1^3}\right)\end{array}$

From Gauss’s law you can write,

$\begin{array}{c}\mathrm{Ï•}=∫Eâ‹\dots dA\\ =\frac{1}{{\mathrm{Î\mu }}_o}q\end{array}$

Substitute known values in the above equation.

$\begin{array}{c}4\mathrm{Ï€}{r}^{2}E=\frac{1}{{\mathrm{Î\mu }}_o}\frac{Q(r^3−r_1^3)}{(r_2^3−r_1^3)}\\ E=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{Q}{r^2}\left(\frac{r^3−r_1^3}{r_2^3−r_1^3}\right)\end{array}$

If is the potential at the outer surface$(r=r_2)$,
Then the potential at point r is given as,

$$\begin{gathered} V=V_s−\underset{r_2}{\overset{r}{}}Eâ‹\dots dr \\ \begin{array}{c}V=V_s−\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o(r_2^3−r_1^3)}\underset{r_2}{\overset{r}{}}(r−\frac{r_1^3}{r^2})dr\\ =V_s−\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o(r_2^3−r_1^3)}\left(\frac{r^2}{2}−\frac{r_2^2}{2}+\frac{r_1^2}{r}+\frac{r_1^3}{r_2}\right)\end{array} \end{gathered}$$

Now, at $r=r_2$, $V_s=\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{r}_2}$

Therefore, the electric potential will be,

$V=\frac{Q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o(r_2^3−r_1^3)}\left(\frac{3r_2^2}{2}−\frac{r^2}{2}−\frac{r_1^3}{r}\right)$

But as known that, the volume charge density is,

$\mathrm{ÒÏ}=\frac{3Q}{4\mathrm{Ï€}}(r_2^3−r_1^3)$

Thus,the electric potential is,

$V=\left(\frac{\mathrm{ÒÏ}}{3{\mathrm{Î\mu }}_o}\right)\left(1.5r_2^2−0.5r^2−r_1^3{r}^{−1}\right)$

Hence, the electric potential at $r_2>r>r_1$ is $V=\left(\frac{\mathrm{ÒÏ}}{3{\mathrm{Î\mu }}_o}\right)(1.5r_2^2−0.5r^2−r_1^3{r}^{−1})$.

If radius$r<r_1$, the point lies inside of the shell.So, the electric potential must be the same.

If$r_1=r$in part (b),

The value of V is given as:

$V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}â‹\dots \frac{3Q}{2}\frac{r_2^2−r_1^2}{(r_2^3−r_1^2)}$

$V=\left(\frac{\mathrm{ÒÏ}}{2{\mathrm{Î\mu }}_o}\right)(r_2^2−r_1^2)$

Hence,the electric potential at $r<r_1$ is $V=\left(\frac{\mathrm{ÒÏ}}{2{\mathrm{Î\mu }}_o}\right)(r_2^2−r_1^2)$.

If$r=r_2$and$r=r_1$, the solutions for electric potential are equal.

Hence, the electric potential at $r=r_2$and$r=r_i$ are equal.