Question

A uniformly charged conducting sphere of1.2 mdiameter has surface charge density 8.1 mC/m² . Find (a) the net charge on the sphere and (b) the total electric flux leaving the surface.

Short answer

1. The net charge on the sphere is $3.7\times {10}^{-5}\mathrm{C}$.
2. The total electric flux leaving the surface is $4.1\times {10}^6\mathrm{N}·{\mathrm{m}}^2/\mathrm{C}$.

Step-by-step solution

The given data

1. Diameter of the sphere,$\mathrm{D}=1.2\mathrm{m}$
2. Surface charge density, $\mathrm{\sigma }=8.1\mathrm{\mu C}/{\mathrm{m}}^2$

Understanding the concept of Gauss law-planar symmetry

Using the concept of surface charge density, we can get the charge accumulated by the sphere. Again using this charge value, we can get the electric flux leaving the surface by using the concept of the Gauss flux theorem.

Formulae:

The surface charge density,

$\mathrm{\sigma }=\frac{\mathrm{q}}{\mathrm{A}}$ (1)

The electric flux leaving the surface,

$\mathrm{\varphi }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ (2)

a) Calculation of the net charge

The surface area of the sphere:

$$\begin{gathered} \mathrm{A}=\left(4{\mathrm{\pi R}}^2\right) \\ ={\mathrm{\pi D}}^2 \end{gathered}$$.

Now, using this value in equation (1), we can get the net charge on the sphere as:

$$\begin{gathered} \mathrm{q}=\mathrm{\sigma }\left({\mathrm{\pi D}}^2\right) \\ =\mathrm{\pi }{\left(1.2\mathrm{m}\right)}^2\left(8.1\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2\right) \\ =3.7\times {10}^{-5}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $3.7\times {10}^{-5}\mathrm{C}$.

b) Calculation of the electric flux

Using the charge value in equation (2), we can get the electric flux leaving the surface of the sphere as:

$$\begin{gathered} \mathrm{\varphi }=\frac{3.7\times {10}^{-5}\mathrm{C}}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2} \\ =4.1\times {10}^6\mathrm{N}·{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$.

Hence, the value of the electric flux is $4.1\times {10}^6\mathrm{N}·{\mathrm{m}}^2/\mathrm{C}$.