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Chapter 3: Problem 29

Compressing a sphere ** A spherical conducting shell has radius \(R\) and potential \(\phi\). If you want, you can consider it to be part of a capacitor with the other shell at infinity. You compress the shell down to essentially zero size (always keeping it spherical) while a battery holds the potential constant at \(\phi\). By calculating the initial and final energies stored in the system, and also the work done by (or on) you and the battery, verify that energy is conserved. (Be sure to specify clearly what your conservation-of-energy statement is, paying careful attention to the signs of the various quantities.)

Short Answer

Expert verified
The work done by you and the battery to compress the shell is \(-\frac{1}{2} (4 \pi \epsilon R) \phi^2\), implying energy is conserved.

Step by step solution

01

Initial Energy

The initial energy stored in the system is associated with the stored electric potential energy, given by \(U = \frac{1}{2} C \phi^2\), where \(C\) is the capacitance of the spherical capacitor. For a spherical capacitor, the capacitance is given by \(C = 4 \pi \epsilon R\). Therefore, the initial energy is \(U_i = \frac{1}{2} (4 \pi \epsilon R) \phi^2\).
02

Final Energy

The shell is compressed to essentially zero size. Therefore, the radius R becomes 0, but the potential \(\phi\) is held constant. Following the same formula as in Step 1, \(U = \frac{1}{2} C \phi^2\), the capacitance would now be \(C = 4 \pi \epsilon (0)\) which results in zero capacitance. Therefore, the final energy is \(U_f = 0\).
03

Conservation of Energy

Conservation of energy requires the sum of the initial and final energy and the work done by you or the battery to be equal. The work done is considered as negative because we compress the shell (doing work on the system). Therefore, from conservation of energy, we have \(U_i + W = U_f\). The work done by you and the battery is \(W = U_f - U_i\). When substituting \(U_f\) as 0 and \(U_i\) as \(\frac{1}{2} (4 \pi \epsilon R) \phi^2\) we have \(W = -\frac{1}{2} (4 \pi \epsilon R) \phi^2\). This confirms that energy is conserved.

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Key Concepts

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

Spherical Conductor
A spherical conductor is a three-dimensional shape where the surface is perfectly symmetrical from its center. This symmetry makes it an interesting object in electrostatics, particularly useful in capacitor studies. A spherical conductor can store electrical charges uniformly on its surface, due to its shape and the properties of conductors.
  • In a spherical conductor, the electric field is zero inside the shell because charges reside on the surface.
  • Due to its symmetry, it is easier to calculate potential functions and capacitance for spherical conductors.
The characteristic property of a spherical conductor is its capacitance, which is related to its size (or radius) and the ability to store charge. The outlined formula for its capacitance is important for solving problems related to energy storage and transfer in electrostatic systems.
Electric Potential Energy
Electric potential energy is the energy that a system of charges possesses due to their positions. In the context of a capacitor, this energy relates to the stored electrical potential within the capacitor's electric field. The key equation for describing this energy is given by:\[ U = \frac{1}{2} C \phi^2 \]Where:
  • \( U \) is the electric potential energy.
  • \( C \) is the capacitance, which in a spherical component is defined by \( C = 4 \pi \epsilon R \).
  • \( \phi \) is the potential.
When you compress the spherical shell, the stored energy changes because the capacitance changes. The formula above shows how energy depends on both the capacitance and the square of the voltage.
Before compression, energy is stored due to the potential across the sphere. Afterward, if the radius is zero, the capacitance becomes zero, eliminating the stored energy. This dependence underscores how changes in the physical parameters of a capacitor directly affect its energy profile.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In the context of compressing a spherical conductor, this principle helps verify the steps involved in energy transformation.
Initially, the energy is stored in the electric field across the spherical shell. During compression, when the radius goes to zero, that energy must go somewhere - it doesn't just disappear. It is either transferred as work done by or on an external agent (like a battery) or stored in a different form.
The equation we use to express conservation of energy in this scenario is:\[ U_i + W = U_f \]Where:
  • \( U_i \) is the initial energy.
  • \( U_f \) is the final energy, which is zero when the sphere is compressed to zero radius.
  • \( W \) is the work done during the compression.
Thus, it is essential to account for the work done, which balances the differences in initial and final energies, guaranteeing that the system obeys the conservation of energy. The initial energy stored which becomes zero after compression signifies that the work done is equal to the negative of the initial stored energy.

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Most popular questions from this chapter

Conductor in a capacitor (a) The plates of a capacitor have area \(A\) and separation \(s\) (assumed to be small). The plates are isolated, so the charges on them remain constant; the charge densities are \(\pm \sigma .\) A neutral conducting slab with the same area \(A\) but thickness \(s / 2\) is initially held outside the capacitor, see Fig. \(3.40 .\) The slab is released. What is its kinetic energy at the moment it is completely inside the capacitor? (The slab will indeed get drawn into the capacitor, as evidenced by the fact that the kinetic energy you calculate will be positive.) (b) Same question, but now let the plates be connected to a battery that maintains a constant potential difference. The charge densities are initially \(\pm \sigma\). (Don't forget to include the work done by the battery, which you will find to be nonzero.)

Transformations of \(\lambda\) and \(I\), Consider a composite line charge consisting of several kinds of carriers, each with its own velocity. For each kind, labeled by \(k\), the linear density of charge measured in frame \(F\) is \(\lambda_{k}\), and the velocity is \(\beta_{k} c\) parallel to the line. The contribution of these carriers to the current in \(F\) is then \(I_{k}=\lambda_{k} \beta_{k} c\). How much do these \(k\)-type carriers contribute to the charge and current in a frame \(F^{\prime}\) that is moving parallel to the line at velocity \(-\beta c\) with respect to \(F ?\) By following the steps we took in the transformations in Fig. \(5.22\), you should be able to show that $$ \lambda_{k}^{\prime}=\gamma\left(\lambda_{k}+\frac{\beta I_{k}}{c}\right), \quad I_{k}^{\prime}=\gamma\left(I_{k}+\beta c \lambda_{k}\right) $$ If each component of the linear charge density and current transforms in this way, then so must the total \(\lambda\) and \(I\) : $$ \lambda^{\prime}=\gamma\left(\lambda+\frac{\beta I}{c}\right), \quad I^{\prime}=\gamma(I+\beta c \lambda) $$ You have now derived the Lorentz transformation to a parallelmoving frame for any line charge and current, whatever its composition,

Principal radii of curvature * Consider a point on the surface of a conductor. The principal radii of curvature of the surface at that point are defined to be the largest and smallest radii of curvature there. To find the radii of curvature, consider a plane that contains the normal to the surface at the given point. Rotate this plane around the normal, and look at the curve representing the intersection of the plane and the surface. The radius of curvature is defined to be the radius of the circle that locally matches up with the curve. For example, a sphere has its principal radii everywhere equal to the radius \(R\). A cylinder has tone principal radius equal to the cross-sectional radius \(R\), and the other equal to infinity. It turns out that the spatial derivative (in the direction of the\\} normal) of the electric field just outside a conductor can be written in terms of the principal radii, \(R_{1}\) and \(R_{2}\), as follows: $$ \frac{d E}{d x}=-\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) E $$ (a) Verify this expression for a sphere, a cylinder, and a plane. (b) Prove this expression. Use Gauss's law with a wisely chosen pillbox just outside the surface. Remember that near the surface, the electric field is normal to it.

Resistances in a cube ** A cube has a resistor \(R\) along each edge. Find the equivalent resistance between two nodes that correspond to: (a) diagonally opposite corners of the cube; (b) diagonally opposite corners of a face; (c) adjacent corners. You do not need to solve a number of simultaneous equations; instead use symmetry arguments. Hint: If two vertices are at the same potential, they can be collapsed to one point without changing the equivalent resistance between the two given nodes.

Two ways of calculating energy \(* *\) * A capacitor consists of two arbitrarily shaped conducting shells, with one inside the other. The inner conductor has charge \(Q\), the outer has charge \(-Q\). We know of two ways of calculating the energy \(U\) stored in this system. We can find the electric field \(E\) and then integrate \(\epsilon_{0} E^{2} / 2\) over the volume between the conductors. Or. if we know the potential difference \(\phi\), we can write \(U=Q \phi / 2\) (or equivalently \(U=C \phi^{2} / 2\) ). (a) Show that these two methods give the same energy in the case of two concentric shells. (b) By using the identity \(\nabla \cdot(\phi \nabla \phi)=(\nabla \phi)^{2}+\phi \nabla^{2} \phi\), show that the two methods give the same energy for conductors of any shape.
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