Question

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?

Short answer

In conclusion, the total electrostatic potential energy of Earth is approximately \(2.05\times10^{12}\ \mathrm{J}\), while the gravitational potential energy is around \(-2.24\times10^{32}\ \mathrm{J}\). The ratio of the two values is about \(-9.2\times10^{-21}\), which means that electrostatic forces have a negligible effect on Earth's structure compared to the gravitational forces.

Step-by-step solution

a) Total Electrostatic Potential Energy

First, we need to calculate the total electric charge on Earth's surface. The relationship between the electric field (E) and the surface charge density (σ) is given by the equation: \(E = \frac{\sigma}{\epsilon_0}\) where \(\epsilon_0 = 8.854\times10^{-12} \mathrm{~C^2/N\cdot m^2}\) is the vacuum permittivity. Now, rearrange the equation to solve for σ: \(\sigma = \epsilon_0 \times E\) Plug in the given electric field value, \(E = -150 \frac{\mathrm{V}}{\mathrm{m}}\): \(\sigma = (8.854\times10^{-12} \mathrm{~C^2/N\cdot m^2}) \times (-150 \frac{\mathrm{V}}{\mathrm{m}})\) \(\sigma \approx -1.33\times10^{-9} \mathrm{~C/m^2}\) Next, we need to calculate the total surface charge by multiplying the charge density by the surface area (A) of the sphere: \(Q = \sigma \times A\) The total surface area of a sphere can be computed using the following formula: \(A = 4 \pi r^2\) Where r is the radius of the sphere. Given the radius of the Earth is \(6,371\times10^{3}\ \mathrm{m}\), the surface area A can be determined: \(A \approx 5.1\times10^{14}\ \mathrm{m^2}\) Now, we can calculate the total surface charge Q: \(Q = (-1.33\times10^{-9} \mathrm{~C/m^2})\times(5.1\times10^{14}\ \mathrm{m^2})\) \(Q \approx -7\times10^5\ \mathrm{C}\) Finally, the total electrostatic potential energy (U) can be calculated using the following formula: \(U = \frac{Q^2}{8\pi\epsilon_0 R}\) Plug in the values for Q, \(\epsilon_0\), and R: \(U \approx \frac{(-7\times10^5\ \mathrm{C})^2}{8\pi(8.854\times10^{-12}\ \mathrm{C^2/N\cdot m^2})(6.371\times10^{3}\ \mathrm{m})}\) \(U \approx 2.05\times10^{12}\ \mathrm{J}\)

b) Gravitational Potential Energy

Next, we need to calculate the gravitational potential energy (GPE) of the Earth treated as a uniform solid sphere. The GPE can be calculated using the following formula: \(GPE = - \frac{3}{5}\frac{GM^2}{R}\) Where G is the gravitational constant (\(6.674\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}\)), M is the mass of the Earth (\(5.972\times10^{24}\ \mathrm{kg}\)), and R is the radius of the Earth (\(6.371\times10^{3}\ \mathrm{m}\)). Plug in these values into the GPE formula: \(GPE \approx - \frac{3}{5}\frac{(6.674\times10^{-11}\ \mathrm{N\cdot m^2/kg^2})(5.972\times10^{24}\ \mathrm{kg})^2}{(6.371\times10^{3}\ \mathrm{m})}\) \(GPE \approx -2.24\times10^{32}\ \mathrm{J}\)

c) Comparison of Electrostatic and Gravitational Forces

Now that we have calculated the total electrostatic potential energy (U) and the gravitational potential energy (GPE) of the Earth, we can compare the two values to determine the extent to which electrostatic forces affect the Earth's structure. The ratio of the electrostatic potential energy to the gravitational potential energy can be computed as follows: \(Ratio = \frac{U}{GPE}\) \(Ratio \approx \frac{2.05\times10^{12}\ \mathrm{J}}{-2.24\times10^{32}\ \mathrm{J}}\) \(Ratio \approx -9.2\times10^{-21}\) The ratio of the electrostatic potential energy to the gravitational potential energy is very small, at approximately \(-9.2\times10^{-21}\). This indicates that electrostatic forces play an insignificant role in the Earth's structure compared to gravitational forces.

Key concepts

Understanding Electric Fields

One cannot discuss electrostatic potential energy without understanding electric fields. An electric field, denoted as \( \vec{E} \), is a vector field that represents the force experienced by a positive test charge placed within the influence of other static charges. The strength and direction of this field are crucial as they determine how charged bodies interact with each other.

The Earth itself generates an electric field that impacts charges on its surface and within its atmosphere. This invisible field is essential when computing the electrostatic potential energy, which is the energy a charge possesses due to its position within an electric field. The given problem considers the Earth as a giant conductor with a field of \( -150 \frac{\mathrm{V}}{\mathrm{m}} \) at its surface. This negative sign indicates the direction of the field towards the center of the Earth, a detail that shapes the analysis of potential energy.

To make sense of the step-by-step solution provided, it's helpful to grasp how surface charge density, \( \sigma \), interacts with the electric field. The formula \( E = \frac{\sigma}{\epsilon_0} \) is a simplified form of Gauss's law for a planar charge, which relates the magnitude of the electric field directly to the amount of charge residing on the surface. So, the stronger the electric field, the greater the surface charge density, and vice versa.

• An electric field is a representation of the force per unit charge.
• The Earth's surface electric field influences charges and can be used to calculate potential energy.
• The relation between surface charge density and electric field is derived from fundamental electrostatic principles like Gauss's law.

Surface Charge Density: A Closer Look

Surface charge density, denoted by \( \sigma \), is a measure of the amount of electric charge per unit area on a surface. For instance, the Earth's charge creates an electric field, and understanding surface charge density helps us compute the total charge on the Earth's surface, necessary for finding the electrostatic potential energy.

In this particular exercise, the provided electric field on Earth's surface and the permittivity of free space are used to calculate \( \sigma \), which is a crucial step in determining the total electric charge. This value is vital because the electrostatic potential energy is dependent on the total charge \( Q \) one is dealing with.

To calculate \( Q \) from \( \sigma \), one must multiply \( \sigma \) by the surface area over which the charge is spread. In this case, Earth's spherical shape means using the formula for a sphere's surface area, \(4 \pi r^2\), with Earth's radius to get the total area, and thus, the total charge.

• Surface charge density is the electric charge per unit of area.
• Computing \( \sigma \) is a critical step in finding the total charge, leading to the calculation of electrostatic potential energy.
• The total electric charge is obtained by multiplying the surface charge density by the surface area.

Gravitational Potential Energy in the Context of Earth

While electrostatic potential energy is about charge interactions, gravitational potential energy (GPE) relates to mass and the gravitational force. Specifically, it is the energy an object possesses due to its position in a gravitational field. For massive bodies like Earth, the GPE is substantial due to Earth's signifcant mass.

In our exercise's solution, the GPE of the Earth is calculated assuming it is a uniform solid sphere. This is a simplified model that allows us to use an equation derived from the law of universal gravitation. The negative sign in the formula accounts for the fact that work must be done against the gravitational field to separate the masses.

Comparing GPE with electrostatic potential energy provides insights into the relative influence of these two fundamental forces on Earth's structure. Our exercise shows that gravitational forces overwhelmingly dominate, evidenced by the calculated ratio of electrostatic to gravitational potential energy. Thus, gravitational forces are principally responsible for Earth's structural integrity, while electrostatic forces are negligible in comparison.

• Gravitational potential energy depends on an object's mass and position within a gravitational field.
• Earth's GPE is calculated by considering the planet as a uniform solid sphere with a specific formula, highlighting the importance of mass and radius.
• The negligible ratio of electrostatic to gravitational potential energy signifies the dominance of gravity over electrostatic forces in shaping Earth's structure.