Question

Charge of uniform volume density $\mathbf{\rho }\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\hspace{0.33em}}\mathbf{\mu C}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ fills a nonconducting solid sphere of radius$\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{.}$ . What is the magnitude of the electric field

(a) $\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}\mathbf{}$and

(b) $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{}$from the sphere’s center?

Short answer

a) The magnitude of the electric fieldat3.5 cm from the center of the sphereis${\mathrm{4.2\times 10}}^3\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

b) The magnitude of the electric field at 8.0 cm from the center of the sphere is$2.4\times {10}^3\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

Step-by-step solution

Listing the given quantities

• Volume Charge density$\mathrm{\rho }=3.2\hspace{0.33em}\mathrm{\mu C}/{\mathrm{m}}^3$
• radius of the sphere = 5.0 cm

Understanding the concept of charge density and electric field

Since the charge distribution is uniform, we can find the total charge q by multiplying by the spherical volume$\left(\frac{4}{3}{\mathrm{\pi r}}^3\right)$with r = R = 0.050 m.

This gives q = 1.68 nC.

Step 3: (a) Calculations for the magnitude of the electric field at r = 3.5 cm from the center of the sphere

Given r = 0.035 m.

The electric field can be calculated as:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{internal}}=\frac{\left|\mathrm{q}\right|\mathrm{r}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3} \\ =\frac{\left(9.0\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(1.68\times {10}^{-9}\mathrm{C}\right)\left(0.035\mathrm{m}\right)}{{\left(0.050\mathrm{m}\right)}^3} \\ =4.2\times {10}^3\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field at 3.5 cm from the center of the sphere is$4.2\times {10}^3\mathrm{N}/\mathrm{C}$ .

(b) Calculations for the magnitude of the electric field at r=8.0 cm from the center of the sphere

Outside the sphere, we have (with r = 0.080 m), then the electric field can be calculated as:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{internal}}=\frac{\left|\mathrm{q}\right|\mathrm{r}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \\ =\frac{\left(9.0\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(1.68\times {10}^{-9}\mathrm{C}\right)}{{\left(0.080\mathrm{m}\right)}^2} \\ =2.4\times {10}^3\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field at 8.0 cm from the center of the sphere is $2.4\times {10}^3\mathrm{N}/\mathrm{C}$.