Question

A uniformly charged conducting sphere of $2.4\mathrm{m}$ diameter has a surface density of $80.0\mathrm{\mu C}/{\mathrm{m}}^2$. (a) Find the charge on the sphere (b) What is the total electric flux leaving the surface of the sphere?

Answer 1

Step 1: Given information

Diameter of the sphere, $\mathrm{d}=2.4\mathrm{m}$

The radius of the sphere, $\mathrm{r}=1.2\mathrm{m}$

Surface charge density, $$\begin{gathered} \mathrm{\sigma }=80.0\mathrm{\mu C}/{\mathrm{m}}^2 \\ =80\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$

Step 2: Finding the charge of the sphere

We know,

The total charge on the surface of the sphere,

$$\begin{gathered} \mathrm{Q}=\mathrm{Charge}\mathrm{density}\times \mathrm{Surface}\mathrm{area} \\ =\mathrm{\sigma }\times 4{\mathrm{\pi r}}^2 \end{gathered}$$

By substituting the values, we get

$$\begin{gathered} =80\times {10}^{-6}\times 4\times 3.14\times {(1.2)}^2 \\ =1.447\times {10}^{-3}\mathrm{C} \end{gathered}$$

Step 3: Finding the total electric flux leaving the surface of the sphere

We know the relation,

${\mathrm{\varphi }}_{\mathrm{total}}=\frac{\mathrm{Q}}{{\mathrm{\epsilon }}_0}$

Permittivity of free space, ${\mathrm{\epsilon }}_0=8.854\times {10}^{-12}{\mathrm{N}}^{-1}{\mathrm{C}}^2{\mathrm{m}}^{-2}$

Therefore,

$$\begin{gathered} {\mathrm{\varphi }}_{\mathrm{total}}=\frac{1.447\times {10}^{-3}}{8.854\times {10}^{-12}} \\ =1.63\times {10}^8\mathrm{N}{\mathrm{C}}^{-1}{\mathrm{m}}^2 \end{gathered}$$