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Energy of concentric shells * (a) Concentric spherical shells of radius \(a\) and \(b\), with \(a

Short Answer

Expert verified
The energy stored in the electric field of the two concentric spherical shells, whether transferred immediately or gradually, is given by \(\frac{Q^2}{8\pi\epsilon_{0}a}\)

Step by step solution

01

Calculation of stored energy for part (a)

The energy stored in the system of two charged spheres can be given by the formula \(U = \frac{1}{2}QV\), where \(Q\) is the charge and \(V\) is the potential difference between the two spheres(same as potential of the inner sphere due to its own charge because potential of the outer sphere due to its own charge is zero at the location of the inner sphere). The potential \(V\) for a charged sphere is given by \(V = \frac{Q}{4\pi\epsilon_{0}a}\), where \(\epsilon_{0}\) is the permittivity of free space. Substituting and simplifying we get \(U = \frac{Q^2}{8\pi\epsilon_{0}a}\)
02

Calculation of work for part (b)

The work done \(dw\) in transferring a small charge \(dq\) from the outer shell to the inner shell is equal to the product of the small charge and the potential \(V(q)\) at that moment. Hence, \(dw = dq * V(q)\). Now we need to integrate this to find the total work done or total energy transferred. \(\int dw = \int dq * V(q)\), from \(q=0\) to \(Q\). This gives us the total amount of energy or work done in moving the charge bit by bit, where \(V(q) = \frac{q}{4\pi\epsilon_{0}a}\). After performing the integral, we discover that the result is the same as in the part (a), \(U = \frac{Q^2}{8\pi\epsilon_{0}a}\)
03

Conclusion

Therefore, the energy stored in the electric field of the system for both case (a) and case (b) is found to be the same: \(\frac{Q^2}{8\pi\epsilon_{0}a}\). This shows that the energy stored in a two concentric spherical shell system only depends on the charge and size of the system, not on how the charge was transferred or distributed between the shells.

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Key Concepts

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

Concentric Spherical Shells
Concentric spherical shells are spherical structures where one sphere is placed inside another without touching each other. Imagine two perfectly round balls rearranged, one inside the other. Each shell can carry a charge, and in our context, the inner shell carries a charge \(Q\), while the outer shell carries a charge \(-Q\). These charges are uniformly distributed over their respective surfaces.

Due to their spherical symmetry, the behavior of charge and the electric field between these shells can be predicted using the principles of electrostatics. The electric field is generally concentrated in the space between the shells and is affected by the charges present. This system is a great way to model various phenomena in electrostatics, helping students understand the importance of symmetry in simplifying complex electric fields.

The setup of concentric spherical shells can be seen in many practical applications, such as capacitors and insulation systems, highlighting their role in the safe and efficient management of electric fields.
Electrostatics
Electrostatics is the branch of physics that deals with the properties and behavior of static electricity. It focuses on forces, fields, and energy associated with stationary or slow-moving electric charges. In the context of concentric spherical shells, electrostatics is essential to determine how charges distribute themselves and how the electric field forms between the shells.

One fundamental concept in electrostatics is Coulomb's Law, which describes the force between two charges. Another key aspect is the electric field, represented as the force per unit charge. For a spherically symmetric charge distribution, like in concentric shells, it simplifies calculations of the field and potential.

Understanding electrostatics is vital for predicting how charges interact in static systems. This includes calculating the potential energy between charged objects, enabling us to determine the energy stored in electric fields, a crucial factor to consider in designs such as capacitors and electrical insulation.
Potential Energy
Potential energy in the context of electrostatics is the energy a charged object possesses due to its position in an electric field. When dealing with concentric spherical shells, it’s important to calculate the potential energy to understand the energy stored within the system's electric field.

The potential energy due to electric forces is derived from the work done in bringing a charge from infinity to a specific point in space. In the scenario of spherical shells, you can calculate the stored energy using the expression \(U = \frac{1}{2} QV\), where \(Q\) is the charge and \(V\) is the potential difference induced by these charges.

By understanding potential energy, students learn how energy is conserved and transferred within electric fields. This concept simplifies the analysis of complex electrostatic systems by providing a scalar quantity, thereby avoiding vector calculations of electric forces over the entire space.
Permittivity of Free Space
The permittivity of free space, denoted by \(\epsilon_{0}\), is a fundamental constant in physics that characterizes the ability of the classical vacuum to support electric fields. It's crucial in the study of electrostatics and affects how electric fields behave between charged objects like concentric spherical shells.

The value of \(\epsilon_{0}\) is approximately \(8.85 \times 10^{-12}\) F/m (farads per meter), and it features prominently in equations like Coulomb's law and the formulae for electric field calculations, such as in our concentric shells' potential \(V = \frac{Q}{4\pi\epsilon_{0}a}\).

In practice, the permittivity of free space helps define how much resistance a vacuum offers to the electric field. It's essential when calculating potential difference and produced energy in systems involving electrical charges. Understanding \(\epsilon_{0}\) gives students insight into the factors that determine how electric fields exist and propagate in empty space, providing a deeper view into fundamental electrostatic principles.

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Most popular questions from this chapter

Flux through a cube (a) A point charge \(q\) is located at the center of a cube of edge \(d\). What is the value of \(\int \mathbf{E} \cdot d \mathbf{a}\) over one face of the cube? (b) The charge \(q\) is moved to one corner of the cube. Now what is the value of the flux of \(\mathbf{E}\) through each of the faces of the cube? (To make things well defined, treat the charge like a tiny sphere.)

Gravity vs. electricity (a) In the domain of elementary particles, a natural unit of mass is the mass of a nucleon, that is, a proton or a neutron, the basic massive building blocks of ordinary matter. Given the nucleon mass as \(1.67 \cdot 10^{-27} \mathrm{~kg}\) and the gravitational constant G as \(6.67 \cdot 10^{-11} \mathrm{~m}^{3} /\left(\mathrm{kg} \mathrm{s}^{2}\right)\), compare the gravitational attraction of two protons with their electrostatic repulsion. This shows why we call gravitation a very weak force. (b) The distance between the two protons in the helium nucleus could be at one instant as much as \(10^{-15} \mathrm{~m}\). How large is the force of electrical repulsion between two protons at that distance? Express it in newtons, and in pounds. Even stronger is the nuclear force that acts between any pair of hadrons (including neutrons and protons) when they are that close together.

(a) Two rings with radius \(r\) have charge \(Q\) and \(-Q\) uniformly distributed around them. The rings are parallel and located a distance \(h\) apart, as shown in Fig. \(1.35\). Let \(z\) be the vertical coordinate, with \(z=0\) taken to be at the center of the lower ring. As a function of \(z\), what is the electric field at points on the axis of the rings? (b) You should find that the electric field is an even function with respect to the \(z=h / 2\) point midway between the rings. This implies that, at this point, the field has a local extremum as a function of \(z\). The field is therefore fairly uniform there; there are no variations to first order in the distance along the axis from the midpoint. What should \(r\) be in terms of \(h\) so that the field is very uniform? By "very" uniform we mean that additionally there aren't any variations to second order in \(z\). That is, the second derivative vanishes. This then implies that the leading-order change is fourth order in \(z\) (because there are no variations at any odd order, since the field is an even function around the midpoint). Feel free to calculate the derivatives with a computer.

Field in the end face Consider a half-infinite hollow cylindrical shell (that is, one that extends to infinity in one direction) with uniform surface charge density. Show that at all points in the circular end face, the electric field is parallel to the cylinder's axis. Hint: Use superposition, along with what you know about the field from an infinite (in both directions) hollow cylinder.

Thundercloud You observe that the passage of a particular thundercloud overhead causes the vertical electric field strength in the atmosphere, measured at the ground, to rise to \(3000 \mathrm{~N} / \mathrm{C}\) (or \(\mathrm{V} / \mathrm{m})\). (a) How much charge does the thundercloud contain, in coulombs per square meter of horizontal area? Assume that the width of the cloud is large compared with the height above the ground. (b) Suppose there is enough water in the thundercloud in the form of \(1 \mathrm{~mm}\) diameter drops to make \(0.25 \mathrm{~cm}\) of rainfall, and that it is those drops that carry the charge. How large is the electric field strength at the surface of one of the drops?

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