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A \(20.0-\mu \mathrm{F}\) spherical capacitor is composed of two concentric metal spheres, one having a radius twice as large as the other. The region between the spheres is a vacuum. Determine the volume of this region.

Short Answer

Expert verified
The volume of the region between the spheres is given by \(V = \frac{4}{3}\pi (2r)^3 - \frac{4}{3}\pi r^3\).

Step by step solution

01

Calculate the Volume of the Outer Sphere

Let's represent the radius of the outer sphere as \(r1\). Since we are given that the radius of the outer sphere is twice as large as the other, we can say that \(r1 = 2r\), where 'r' is the radius of the inner sphere. The volume of the outer sphere \(V1\) can be calculated using the formula \(V1 = \frac{4}{3}\pi r1^3\). By substituting \(r1 = 2r\), we get \(V1 = \frac{4}{3}\pi (2r)^3\).
02

Calculate the Volume of the Inner Sphere

For the inner sphere, we use the formula for the volume of a sphere, \(V = \frac{4}{3}\pi r^3\), where 'r' is the radius of the inner sphere. So, the volume of the inner sphere, \(V2\), is calculated as \(V2 = \frac{4}{3}\pi r^3\).
03

Calculate the Volume of the Region Between the Spheres

To find the volume of the region between the spheres, subtract the volume of the inner sphere from the volume of the outer sphere. This can be represented as \(V = V1 - V2 = \frac{4}{3}\pi (2r)^3 - \frac{4}{3}\pi r^3\). Simplify the right hand side to find the volume 'V'.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance
Capacitance is a fundamental concept in electronics and physics that helps us understand how capacitors work. A capacitor is a device that can store electrical energy in an electric field. Capacitance, denoted by the symbol 'C', is the measure of a capacitor's ability to store charge per unit voltage across its plates.
When we talk about spherical capacitors, like the one in our exercise, we have two concentric spheres that make up the capacitor. The inner sphere acts as one plate, while the outer sphere acts as the other. The capacitance of a spherical capacitor can be calculated using the formula:
  • \[C = \frac{4 \pi \varepsilon_0}{ \frac{1}{r2} - \frac{1}{r1}} \]
Where:
  • \(\varepsilon_0\) is the permittivity of free space.
  • \(r1\) is the radius of the outer sphere.
  • \(r2\) is the radius of the inner sphere.
The capacitance depends on the size and distance between the spheres, as well as the material between them, which in this case is a vacuum. Understanding capacitance is crucial for designing circuits and understanding how a capacitor can influence circuit behavior.
Concentric Spheres
Concentric spheres are a common geometric arrangement in physics, especially when dealing with spherical capacitors. "Concentric" means that the spheres share a common center. This setup is important because it ensures uniform electric fields and potential differences between the spheres.
In our exercise, the spherical capacitor consists of two metallic spheres. The inner sphere has a smaller radius, and the outer sphere encloses it. The fact that the outer sphere's radius is twice that of the inner sphere means that their dimensions significantly impact the capacitor's characteristics. The volume of space between these two spheres is what allows the capacitor to function. This volume is crucial for determining the total capacitance and is derived by subtracting the inner sphere's volume from the outer sphere's volume.
  • The shared center point is essential for ensuring symmetry.
  • Uniformity helps in simplifying calculations of electric fields and potentials.
These concepts make it easier to apply mathematical formulas and understand the working of a spherical capacitor.
Vacuum
A vacuum, in the context of physics and capacitors, refers to a space entirely devoid of matter, including air. It's an important factor in our exercise because the space between the two concentric spheres of the capacitor is a vacuum.
The vacuum plays a crucial role in a spherical capacitor as it affects the electric field and capacitance. The absence of any medium means that the only influence on the electric field is the geometry and charge of the spheres themselves. The permittivity of vacuum, represented by \(\varepsilon_0\), is a constant that factors into the capacitance equation.
  • Vacuum has the lowest permittivity possible, which impacts the storage of electric energy.
  • It simplifies calculations as there are no extra materials influencing the electric field.
Understanding the role of a vacuum helps in appreciating how a spherical capacitor's performance is directly linked to the environment between its plates.

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Most popular questions from this chapter

A 1 -megabit computer memory chip contains many \(60.0-\mathrm{fF}\) capacitors. Fach capacitor has a plate area of \(21.0 \times 10^{-12} \mathrm{m}^{2} .\) Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The order of magnitude of the diameter of an atom is \(10^{-10} \mathrm{m}=0.1 \mathrm{nm} .\) Express the plate separation in nanometers.

Two capacitors, \(C_{1}=25.0 \mu \mathrm{F}\) and \(C_{2}=5.00 \mu \mathrm{F},\) are connected in parallel and charged with a \(100-\mathrm{V}\) power supply. (a) Draw a circuit diagram and calculate the total energy stored in the two capacitors. (b) What If? What potential difference would be required across the same two capacitors connected in series in order that the combination stores the same amount of energy as in (a)? Draw a circuit diagram of this circuit.

Two capacitors, \(C_{1}=5.00 \mu \mathrm{F}\) and \(C_{2}=12.0 \mu \mathrm{F},\) are connected in parallel, and the resulting combination is connected to a \(9.00-\mathrm{V}\) battery. (a) What is the equivalent capacitance of the combination? What are (b) the potential difference across each capacitor and (c) the charge stored on each capacitor?

An isolated capacitor of unknown capacitance has been charged to a potential difference of \(100 \mathrm{V}\). When the charged capacitor is then connected in parallel to an uncharged \(10.0-\mu \mathrm{F}\) capacitor, the potential difference across the combination is \(30.0 \mathrm{V}\). Calculate the unknown capacitance.

When a potential difference of \(150 \mathrm{V}\) is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of \(30.0 \mathrm{nC} / \mathrm{cm}^{2} .\) What is the spacing between the plates?

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