Question

Iaking earth to be a metallic spheres, its capacity will approximately be (A) \(6.4 \times 10^{6} \mathrm{~F}\) (B) \(700 \mathrm{pF}\) (C) \(711 \mu \mathrm{F}\) (D) \(700 \mathrm{pF}\)

Short answer

The capacitance of Earth, considering it to be a metallic sphere, can be calculated using the formula \(C = 4 \pi \varepsilon_0 R\), where \(\varepsilon_0\) is the permittivity of free space and \(R\) is the radius of Earth. Plugging in the values \(\varepsilon_0 \approx 8.85 \times 10^{-12} \mathrm{F/m}\) and \(R \approx 6.37 \times 10^6 \mathrm{m}\), we get \(C \approx 712.49 \times 10^{-6} \mathrm{F}\). Comparing this to the given options, the closest value is (C) \(711 \mu \mathrm{F}\), which is the correct answer.

Step-by-step solution

Insert the values into the formula

Now we will insert the values \(\varepsilon_0\) and \(R\) into the capacitance formula: \[C = 4 \pi (8.85 \times 10^{-12} \mathrm{F/m}) (6.37 \times 10^6 \mathrm{m})\]

Solve for capacitance

To solve for capacitance, we just need to do the arithmetic: \[C = 4 \pi(8.85 \times 10^{-12} \mathrm{F/m}) (6.37 \times 10^6 \mathrm{m})\] \[C = 4 \pi(56.4855 \times 10^{-6} \mathrm{F})\] Next, we multiply the values: \[C = 4 \pi(56.4855 \times 10^{-6} \mathrm{F}) \approx 712.49 \times 10^{-6} \mathrm{F}\]

Choose the correct option

The calculated value of Earth's capacitance is approximately \(712.49 \mu \mathrm{F}\). Comparing the calculated value to the given options, we find that option (C) \(711 \mu \mathrm{F}\) is the closest to our result. Hence, the answer is: (C) \(711 \mu \mathrm{F}\)

Key concepts

Capacitance of Earth

In physics, the concept of capacitance refers to the ability of a system to store electrical charge. When we think of Earth as a gigantic metallic sphere, we can calculate its capacitance using a straightforward formula. Capacitance is measured in farads (F), and the larger the capacitance, the more charge an object can store. In this context, the Earth's capacitance is the measure of how much electrical charge it can hold. To determine Earth's capacitance, we assume it behaves like a metallic sphere because this simplifies calculations.

The capacitance of a sphere is given by the formula:

• \(C = 4 \pi \varepsilon_0 R\)

where \(R\) is the radius of the sphere, and \(\varepsilon_0\) is the permittivity of free space. With Earth's radius approximately \(6.37 \times 10^6\) meters, this formula helps us estimate the charge Earth's surface can theoretically store. By solving with Earth's specific measurements, we find that the capacitance comes close to \(711 \mu \mathrm{F}\), showing the enormous capacity of Earth to hold charge due to its vast size.

Permittivity of Free Space

Permittivity of free space, also known as electric constant or \(\varepsilon_0\), is a fundamental physical constant important in electromagnetism. It describes how electric fields affect and are affected by a vacuum. This constant is critical when calculating capacitance because it modifies how a sphere or any object interacts with electric fields.

In practical terms, \(\varepsilon_0\) quantifies the ability of the vacuum to permit electric field lines. This is analogous to how a vacuum or other material affects the forces between charges. The accepted value of permittivity of free space is:

• \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\)

This means that in a vacuum, electric fields become weaker due to this factor, influencing the calculated capacitance for objects in such conditions. Understanding \(\varepsilon_0\) allows us to comprehend how systems behave in absence of material substance, which is pivotal in fields like electrostatics and circuit design.

Metallic Sphere Capacitance

The concept of capacitance is often explored with simple conductive shapes, like spheres, to make understanding complex systems easier. A metallic sphere's capacitance straightforwardly exemplifies this as it emits a uniform electric field outside its surface, minimizing complexities in calculations.

For a sphere, the capacitance can be calculated without resorting to intricate equations:

• \(C = 4 \pi \varepsilon_0 R\)

Here, \(R\) is the radius, and \(\varepsilon_0\) stands for the permittivity of free space. This formula highlights the linear relationship between the sphere’s size (its radius) and its capacitance, displaying how larger spheres can hold more charge simply due to their geometric properties.

The calculation in the original exercise exemplifies this principle, as Earth's large radius directly results in its substantial capacitance value of approximately \(711 \mu \mathrm{F}\). Such analyses of spherical capacitance are foundational for comprehending how more complex objects, like capacitors in a circuit, store charge.