The volume charge density of a solid nonconducting sphere of radius$\mathit{R}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{60}\mathbf{cm}$ varies with radial distance ras given by $\mathbf{\rho }\mathbf{=}\mathbf{(}\mathbf{14}\mathbf{.}\mathbf{1}\mathbf{pC}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}\mathbf{)}\mathit{r}\mathbf{/}\mathit{R}$. (a) What is the sphere’s total charge? What is the field magnitude E, at(b), (c) $\mathit{r}\mathbf{=}\mathit{R}\mathbf{/}\mathbf{2}\mathbf{.}\mathbf{00}$, and (d) $\mathit{r}\mathbf{=}\mathit{R}$? (e) Graph Eversusr.

radius R = 5.60 cm

$\mathrm{\rho }=(14.1\mathrm{pC}/{\mathrm{m}}^3)\frac{\mathrm{r}}{\mathrm{R}}$

We integrate the volume charge density over the volume and require the result

be equal to the total charge

Formula:

$\int \mathrm{dx}\int \mathrm{dy}\int \mathrm{dz\rho }=4\mathrm{\pi }{\int }_0^{\mathrm{R}}\mathrm{dr}{\mathrm{r}}^2\mathrm{\rho }=\mathrm{Q}$

Substituting the expression $\mathrm{\rho }={\mathrm{\rho }}_{\mathrm{S}}\frac{\mathrm{r}}{\mathrm{R}}$, with ${\mathrm{\rho }}_{\mathrm{s}}=14.1\mathrm{pC}/{\mathrm{m}}^3$, and performing the integration leads to,

$4\mathrm{\pi }\left(\frac{{\mathrm{\rho }}_{\mathrm{s}}}{\mathrm{R}}\right)\left(\frac{{\mathrm{R}}^4}{4}\right)=Q$

$$\begin{gathered} Q={\mathrm{\pi \rho }}_{\mathrm{s}}{\mathrm{R}}^3 \\ =\mathrm{\pi }\left(14.1\mathrm{pC}/{\mathrm{m}}^3\right){\left(0.560\mathrm{m}\right)}^3 \\ =7.78\times {10}^{-15}\mathrm{C} \end{gathered}$$

The total charge of the sphere is$7.78\times {10}^{-15}\mathrm{C}$

At r = 0, the electric field is zero (E = 0) since the enclosed charge is zero.

At a certain point within the sphere, at some distance r from the center, the field is given by Gauss’ law:

$\mathrm{E}=\frac{{\mathrm{q}}_{\mathrm{enc}}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$.

$4\mathrm{\pi }{\int }_0^{\mathrm{R}}\mathrm{dr}{\mathrm{r}}^2\mathrm{\rho }=4\mathrm{\pi }\left(\frac{{\mathrm{\rho }}_{\mathrm{s}}}{\mathrm{R}}\right)\left(\frac{{\mathrm{r}}^4}{4}\right)$

therefore,

$$\begin{gathered} \mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{\pi \rho }}_{\mathrm{s}}{\mathrm{r}}^4}{{\mathrm{Rr}}^2} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{\pi \rho }}_{\mathrm{s}}{\mathrm{r}}^2}{\mathrm{R}} \end{gathered}$$

For r = R/2.00, where R = 5.60 cm, the electric field is

$$\begin{gathered} \mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{{\mathrm{\pi \rho }}_{\mathrm{s}}{\left(\mathrm{R}/2\right)}^2}{\mathrm{R}}\right) \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{{\mathrm{\pi \rho }}_{\mathrm{s}}\mathrm{R}}{4}\right) \\ =\frac{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}\right)\mathrm{\tau \tau }\left(14.{10}^{-12}\mathrm{C}/{\mathrm{m}}^3\right)\left(0.0560\mathrm{m}\right)}{4} \\ =5.58\times {10}^{-3}\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field at r = R/2.00 is $5.58\times {10}^{-3}\mathrm{N}/\mathrm{C}$

For r = R, the electric field is

$$\begin{gathered} \mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{{\mathrm{\pi \rho }}_{\mathrm{s}}{\mathrm{R}}^2}{\mathrm{R}}\right) \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{{\mathrm{\pi \rho }}_{\mathrm{s}}\mathrm{R}}{1}\right) \\ =\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\mathrm{\tau \tau }\left(14.1\times {10}^{-12}\mathrm{C}/{\mathrm{m}}^3\right)\left(0.0560\mathrm{m}\right) \\ =2.23\times {10}^{-2}\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field r = R is $2.23\times {10}^{-2}\mathrm{N}/\mathrm{C}$

The electric field strength as a function of r is depicted below: