91Ó°ÊÓ

1. Diameter of the satellite,$\mathrm{D}=1.3\mathrm{m}$
2. The charge accumulated by the satellite,$\mathrm{q}=2.4×{10}^{-6}\mathrm{C}$

Using the basic concept of the surface charge density, we can get the resulting charge density with the given data. Similarly, using the concept of the electric field of Gauss law of a conducting plate, we can get the electric field required.

Formulae:

The surface charge density of a body, $\mathrm{σ}=\frac{\mathrm{q}}{\mathrm{A}}$ (1)

The electric field of a conducting ball,$\mathrm{E}=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}$ (2)

The area of the sphere may be written as:

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

Thus, the resulting surface density of the satellite can be given using equation (i) such that,

$$\begin{gathered} \mathrm{σ}=\frac{2.4×{10}^{-6}\mathrm{C}}{\mathrm{π}{\left(1.3\mathrm{m}\right)}^2} \\ =4.5×{10}^{-7}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of the charge density is $4.5×{10}^{-7}\mathrm{C}/{\mathrm{m}}^2$.

Using this value of charge density in equation (2), we can get the electric field value as:

$$\begin{gathered} \mathrm{E}=\frac{4.5×{10}^{-7}\mathrm{C}/{\mathrm{m}}^2}{8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2} \\ =5.1×{10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the field is $5.1×{10}^4\mathrm{N}/\mathrm{C}$.