Question

Space vehicles traveling through Earth’s radiation belts can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations. Suppose a spherical metal satellite1.3min diameter accumulates$\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{\mu C}$of charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.

Short answer

1. The resulting surface charge density is $4.7\times {10}^{-7}\mathrm{C}/{\mathrm{m}}^2$.
2. The magnitude of the electric field is $5.1\times {10}^4\mathrm{N}/\mathrm{C}$.

Step-by-step solution

The given data

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

Understanding the concept of the electric field

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

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

a) Calculation of the surface charge density

The area of the sphere may be written as:

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

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

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

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

b) Calculation of the electric field

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

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

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