91Ó°ÊÓ

1. Diameter of the sphere,$\mathrm{D}=1.2\mathrm{m}$
2. Surface charge density, $\mathrm{σ}=8.1\mathrm{μ°ä}/{\mathrm{m}}^2$

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,

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

The electric flux leaving the surface,

$\mathrm{Ï•}=\frac{\mathrm{q}}{{\mathrm{Î\mu }}_0}$ (2)

The surface area of the sphere:

$$\begin{gathered} \mathrm{A}=\left(4{\mathrm{Ï€¸é}}^2\right) \\ ={\mathrm{Ï€\P Ù}}^2 \end{gathered}$$.

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

$$\begin{gathered} \mathrm{q}=\mathrm{σ}\left({\mathrm{Ï€\P Ù}}^2\right) \\ =\mathrm{Ï€}{\left(1.2\mathrm{m}\right)}^2\left(8.1×{10}^{-6}\mathrm{C}/{\mathrm{m}}^2\right) \\ =3.7×{10}^{-5}\mathrm{C} \end{gathered}$$

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

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

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

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