Problem 19 - Fundamentals of Physics

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 satellite \(1.3 \mathrm{~m}\) in diameter accumulates \(2.4 \mu \mathrm{C}\) 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.

Step-by-Step Solution

Answer

(a) The surface charge density is approximately \(4.52 \times 10^{-7} \text{ C/m}^2\). (b) The electric field is about \(51100 \text{ N/C}\).

Step 1: Find the Surface Area of the Satellite

The satellite is a sphere with a diameter of 1.3 m. The radius \(r\) is half of the diameter: \(r = \frac{1.3}{2} = 0.65\) m. The formula for the surface area \(A\) of a sphere is: \[ A = 4\pi r^2 \]Plug in the radius:\[ A = 4\pi (0.65)^2 \approx 5.3093 \text{ m}^2 \]

Step 2: Calculate Surface Charge Density

Surface charge density \(\sigma\) is given by the total charge \(Q\) divided by the surface area \(A\). We have \(Q = 2.4 \times 10^{-6} \text{ C (since } 2.4 \mu\text{C} = 2.4 \times 10^{-6} \text{ C)}\). The formula is:\[ \sigma = \frac{Q}{A} = \frac{2.4 \times 10^{-6} \text{ C}}{5.3093 \text{ m}^2} \approx 4.52 \times 10^{-7} \text{ C/m}^2 \]

Step 3: Determine Electric Field Just Outside the Surface

The electric field \(E\) just outside the surface of a charged sphere can be found using the formula:\[ E = \frac{\sigma}{\varepsilon_0} \]where \(\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2)\) is the permittivity of free space.Substitute the known values:\[ E = \frac{4.52 \times 10^{-7} \text{ C/m}^2}{8.854 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2)} \approx 51100 \text{ N/C} \]

Key Concepts

Surface Charge Density

Surface charge density is a measure of how much electric charge accumulates per unit area on a surface. It is denoted by the symbol \(\sigma\). In this context, we are dealing with a spherical satellite that accumulates charges on its surface. To find the surface charge density, we first need to know two things: the total charge \(Q\) and the surface area \(A\) of the sphere.

For a sphere, the surface area \(A\) can be determined using the formula \(A = 4\pi r^2\), where \(r\) is the radius. Given that the diameter of the satellite is \(1.3\, \text{m}\), the radius \(r\) is half of this value, which is \(0.65\, \text{m}\). Substituting this radius into the formula gives the surface area. Using these numbers, the surface area of the satellite is approximately \(5.3093 \text{ m}^2\).

Given the satellite accumulates a charge of \(2.4 \mu C\) which is equivalent to \(2.4 \times 10^{-6}\, \text{C}\), the surface charge density \(\sigma\) is calculated as \(\sigma = \frac{Q}{A} = \frac{2.4 \times 10^{-6}\, \text{C}}{5.3093\, \text{m}^2}\). This gives a surface charge density of approximately \(4.52 \times 10^{-7}\, \text{C/m}^2\).

• \(\sigma\) is crucial in determining how charges spread over the surface.
• It helps understand the distribution and potential effects of charge accumulations.

Electric Field

The electric field is a vector field surrounding an electric charge that exerts a force on other charges, attracting or repelling them. On the outside of a charged sphere, the electric field \(E\) can be computed using the formula \(E = \frac{\sigma}{\varepsilon_0}\), where \(\sigma\) is the surface charge density and \(\varepsilon_0\) is the permittivity of free space.

In this problem, we previously determined \(\sigma\) to be \(4.52 \times 10^{-7} \text{ C/m}^2\). By applying the formula, you can find the magnitude of the electric field just outside the satellite.

The electric field strength just outside the surface depends on two key elements:

• Surface Charge Density (\(\sigma\)): Higher charge density results in a stronger electric field.
• Permittivity of Free Space (\(\varepsilon_0\)): This constant moderates the relation between the electric field and charge density.

Calculating with \(\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2)\), the electric field is approximately \(51100 \text{ N/C}\). This substantial strength indicates the forces impacting nearby charges. Such a force field is significant in understanding satellite charge leakage and its interaction with ambient particles.

Permittivity of Free Space

The permittivity of free space, often represented as \(\varepsilon_0\), is a fundamental physical constant that describes how electric fields interact with the vacuum of space. It is vital for mathematical expressions of electrostatics, serving as a basis for understanding concepts like electric fields and capacitance.

\(\varepsilon_0\) has a constant value of \(8.854 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2)\). This constant appears in the relations between electric field strength and surface charge density. Specifically, it shows up in equations such as \(E = \frac{\sigma}{\varepsilon_0}\), where it's used to find the electric field strength due to a surface charge density \(\sigma\).

Key roles of \(\varepsilon_0\) include:

• Regulating the force magnitude between charged particles in a vacuum.
• Allowing the calculation of capacitance in configurations like parallel plate capacitors.

Understanding \(\varepsilon_0\) assists in comprehending how charges influence each other even across empty spaces. It helps encapsulate how electric fields propagate, defining the linkage between charge distributions and resultant electrostatics phenomena.