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) Surface charge density is \(4.52 \times 10^{-7} \, \text{C/m}^2\). (b) Electric field magnitude is \(5.11 \times 10^4 \, \text{N/C}\).

Step 1: Find Surface Area of the Satellite

The satellite is a sphere with a diameter of \(1.3\, \text{m}\). To find its surface area, first calculate the radius \(r\), which is half of the diameter: \(r = \frac{1.3}{2} = 0.65\, \text{m}\). Now, use the formula for the surface area of a sphere \(A = 4\pi r^2\):\[A = 4\pi (0.65)^2 = 4\pi \times 0.4225 = 5.31\,\text{m}^2\]

Step 2: Calculate Surface Charge Density

Surface charge density \(\sigma\) is defined as the charge \(Q\) per unit area \(A\). The charge accumulated is \(Q = 2.4 \, \mu \text{C} = 2.4 \times 10^{-6} \, \text{C}\). Using the surface area from Step 1, calculate \(\sigma\):\[\sigma = \frac{Q}{A} = \frac{2.4 \times 10^{-6}}{5.31} = 4.52 \times 10^{-7} \, \text{C/m}^2\]

Step 3: Determine Electric Field Just Outside the Surface

The electric field just outside a charged surface of a conductor is given by Gauss's law: \(E = \frac{\sigma}{\varepsilon_0}\), where \(\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/ ext{N}\cdot\text{m}^2\) is the permittivity of free space. Calculate the electric field:\[E = \frac{4.52 \times 10^{-7}}{8.85 \times 10^{-12}} = 5.11 \times 10^{4} \, \text{N/C}\]

Key Concepts

Electric Field

The concept of the electric field is fundamental in electrostatics. It describes how charged particles, like electrons and protons, interact from a distance. Essentially, an electric field represents the influence a charge exerts on other charges in its vicinity.
A single charge will create a field around it, which decreases in strength as you move further away. The electric field at any point in space is a vector quantity—it has a magnitude and a direction. The direction is defined as the direction that a positive test charge would be pushed or pulled.
In mathematical terms, for a point charge, the electric field is given by the formula:\[ E = \frac{k \, |Q|}{r^2} \]where:

• potential *\( E \)* is the electric field strength
• *\( k \)* represents Coulomb's constant \((8.99 \times 10^9 \, \text{N m}^2/\text{C}^2)\)
• *\( Q \)* is the charge creating the field
• *\( r \)* is the distance from this charge

When considering a conducting surface, like a satellite, a key point is that the electric field just outside is perpendicular to the surface and its magnitude is determined by the surface charge density. This will be elaborated with Gauss's Law later.

Surface Charge Density

Surface charge density, denoted as \( \sigma \), is an important quantity when dealing with charged surfaces in electrostatics. It measures how much charge is spread over a given area of a surface. It's essentially the charge \( Q \) spread over an area \( A \), and mathematically is expressed as:\[ \sigma = \frac{Q}{A} \]where:

• *\( \sigma \)* is the surface charge density
• *\( Q \)* is the total charge on the surface
• *\( A \)* is the area over which the charge is distributed

Understanding surface charge density helps predict the behavior of charged objects. For example, on a satellite, if the charge is concentrated in a small area, the density will be high, increasing the electric field's magnitude just outside that part of the surface. This is crucial for assessing the risk of electrical discharge, which can damage space equipment.

Gauss's Law

Gauss's Law is a cornerstone in understanding electric fields due to symmetric charge distributions. It relates the net electric field flowing out of a closed surface to the charge enclosed within that surface. The law is particularly powerful because it simplifies complex electrostatic problems into more manageable calculations.
The mathematical expression of Gauss's Law is:\[ \Phi = \oint E \cdot dA = \frac{Q_{enc}}{\varepsilon_0} \]where:

• *\( \Phi \)* is the electric flux through a closed surface
• *\( Q_{enc} \)* is the charge enclosed by that surface
• *\( \varepsilon_0 \)* is the permittivity of free space \((8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2)\)

For conductors, Gauss's Law is incredibly helpful. For a uniform surface charge density on a conductor, the electric field just outside the surface is directly proportional to the surface charge density, given by:\[ E = \frac{\sigma}{\varepsilon_0} \]This relationship allows us to compute the electric field near charged bodies like satellites efficiently. Understanding this law can help mitigate risks associated with electronic components in adverse charge environments, like space.