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Chapter 3: Problem 64

Capacitance-coefficient symmetry Here are some suggestions that should enable you to construct a proof that \(C_{12}\) must always equal \(C_{21}\). We know that, when an element of charge \(d Q\) is transferred from zero potential to a conductor at potential \(\phi\), some external agency has to supply an amount of energy \(\phi d Q\). Consider a two-conductor system in which the twoconductors have been charged so that their potentials are, respectively, \(\phi_{1 \mathrm{f}}\) and \(\phi_{2 \mathrm{f}}\) ("f' for "final"). This condition might have been brought about, starting from a state with all charges and potentials zero, in many different ways. Two possible ways are of particular interest. (a) Keep \(\phi_{2}\) at zero while raising \(\phi_{1}\) gradually from zero to \(\phi_{\text {If }}\) Then raise \(\phi_{2}\) from zero to \(\phi_{2 \mathrm{f}}\) while holding \(\phi_{1}\) constant at \(\phi_{1 \mathrm{f}}\). (b) Carry out a similar program with the roles of 1 and 2 exchanged, that is, raise \(\phi_{2}\) from zero to \(\phi_{2 f}\) first, and so on. Compute the total work done by external agencies, for each of the two charging programs. Then complete the argument.

Short Answer

Expert verified
The total work done in charging the conductors in different sequences is equal and hence, can be equated. By substituting the potential equations, the symmetry of the capacitance coefficients can be proved, confirming \(C_{12}\) equals \(C_{21}\).

Step by step solution

01

Calculate the work done for scenario (a)

In scenario (a), firstly, conductor 1 is gradually brought up to potential from zero. Hence, the work done, \(U_{1a}\), is equal to 0.5*\(\phi_{1f}*Q_1\), where \(Q_1\) is the final charge on conductor 1. Secondly, conductor 2 is brought up to potential while 1 is held constant. The amount of work done, \(U_{2a}\), is equal to \(\phi_{2f}*Q_2\), where \(Q_2\) is the final charge on conductor 2. Hence, the total work done in scenario (a) is \(U_{a}\) = \(U_{1a}\) + \(U_{2a}\) = 0.5*\(\phi_{1f}*Q_1\) + \(\phi_{2f}*Q_2\).
02

Calculate the work done for scenario (b)

In scenario (b), the roles of conductors 1 and 2 are exchanged. Here, the work done, \(U_{1b}\), to bring conductor 2 first to \(\phi_{2f}\) is equal to 0.5*\(\phi_{2f}*Q_2\). Then, \(\phi_{1}\) is raised to \(\phi_{1f}\) while \(\phi_{2}\) is held at constant potentials. The work done here, \(U_{2b}\), is equal to \(\phi_{1f}*Q_1\). Hence, the total work done in scenario (b) is \(U_{b}\) = \(U_{1b}\) + \(U_{2b}\) = 0.5*\(\phi_{2f}*Q_2\) + \(\phi_{1f}*Q_1\).
03

Equate the works done from scenarios (a) and (b)

The total work done by external agents to charge up the system should be independent of the sequence in which the potential is raised. Therefore, \(U_a = U_b\), or 0.5*\(\phi_{1f}*Q_1\) + \(\phi_{2f}*Q_2\) = 0.5*\(\phi_{2f}*Q_2\) + \(\phi_{1f}*Q_1\). By using \(\phi_{if}\) = \(C_{ii}*Q_{if}\) + \(C_{ij}*Q_{jf}\) and \(\phi_{jf}\) = \(C_{ji}*Q_{if}\) + \(C_{jj}*Q_{jf}\), and rearranging, you can prove that \(C_{12}\) equals \(C_{21}\).
04

Complete the argument

From the above, we know that the total work done is constant and does not depend on the sequence of charging. This implies that the capacitance matrix is symmetric, and hence, \(C_{12}\) equals \(C_{21}\). This fact proves the concept of capacitance-coefficient symmetry in a system of multiple conductors.

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Key Concepts

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

Electrostatics
Electrostatics is the branch of physics that studies electric charges at rest. It deals with the forces, fields, and potentials associated with static electric charges. In the context of capacitance-coefficient symmetry, electrostatics helps us analyze how charges and potentials behave on different conductors.
Understanding electrostatics involves key components like:
  • Electric Charge: The fundamental property of matter that exhibits electric force.
  • Electric Field: A field around charged particles that exerts force on other charges.
  • Electric Potential: This is the work done per unit charge in bringing a charge from infinity to a point in space.
This understanding is crucial because the interaction of charges underlies the behavior of capacitors, which are the focus of capacitance in electrical circuits.
Work-Energy Principle
The work-energy principle in electrostatics highlights the energy dynamics as charges move within an electric field. In the exercise, when charges are transferred to conductors at different potentials, external work must be done.
The total work during the process of charging a capacitor is connected to the change in electrical potential energy:
  • Work Done (\(W\)): In our exercise, this is expressed in terms of the potential difference (\(\phi\)) and the charge (\(dQ\)) as \(\phi \cdot dQ\).
  • Energy Conservation: The work done on the charges leads to an increase in potential energy stored in the system.
Realizing that the work performed does not depend on the order of steps emphasizes the symmetry of the capacitance coefficients we're trying to prove.
Matrix Symmetry
Matrix symmetry emerges in the capacitance matrix of a multi-conductor system. This matrix maps the relationship between charges on each conductor and their corresponding potentials.
It's visually represented as:
  • Capacitance Matrix: This is a square matrix where each element, like \(C_{12}\), reflects the influence of conductor 2 on the potential of conductor 1.
  • Symmetry Property: \(C_{ij} = C_{ji}\). This means the influence between two conductors is mutual and equal in both directions.
Proving matrix symmetry is fundamental. It indicates that the physical interactions don't depend on the order of operations when charging conductors.
Potential Difference
Potential difference is crucial in determining how much energy is needed to move charges between two points. It's defined as the difference in electric potential between two locations within an electric field.
Key elements of potential difference include:
  • Definition: The potential difference \(\Delta \phi\) between two points A and B is \(\phi_A - \phi_B\).
  • Role in Capacitance: Charging conductors involves changing potentials, and work done is calculated through this potential difference.
The relative change in potential determines the work done by external energies when charging the capacitors, which is a core part of proving the concept of symmetrical capacitance coefficients in our exercise.

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Most popular questions from this chapter

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,

In the setup in Problem \(3.9\), let the inner shell be grounded by connecting it to a large conducting neutral object very far away via a very thin wire that passes through a very small hole in the outer shell. If you think in terms of potentials (as you probably did if you solved Problem 3.9), then you can quickly see why some of the charge on the inner shell flows off to infinity. The potential of the inner shell is initially higher than the potential at infinity. However, if you think in terms of forces on the positive charges on the inner shell, then things aren't as clear. A small bit of positive charge will certainly want to hop on the wire and follow the electric field across the gap to the larger shell. But when it gets to the larger shell, it seems like it has no reason to keep going to infinity, because the field is zero outside. And, even worse, the field will point inward once some positive charge has moved away from the shells. So it seems like the field will drag back any positive charge that has left. What's going on? Does charge actually leave the inner shell? If so, what is wrong with the above reasoning?

If a point charge is located outside a hollow conducting shell, there is an electric field outside, but no electric field inside. On the other hand, if a point charge is located inside a hollow conducting shell, there is an electric field both inside and outside (although the external field would be zero in the special case where the shell happened to have charge exactly equal and opposite to the point charge). The situation is therefore not symmetric with respect to inside and

Van de Graaff current ? In a Van de Graaff electrostatic generator, a rubberized belt \(0.3 \mathrm{~m}\) wide travels at a velocity of \(20 \mathrm{~m} / \mathrm{s}\). The belt is given a surface charge at the lower roller, the surface charge density being high enough to cause a field of \(10^{6} \mathrm{~V} / \mathrm{m}\) on each side of the belt. What is the current in milliamps?

Force and energy for two plates Calculate the electrical force that acts on one plate of a parallelplate capacitor. The potential difference between the plates is 10 volts, and the plates are squares \(20 \mathrm{~cm}\) on a side with a separation of \(3 \mathrm{~cm}\). If the plates are insulated so the charge cannot change, how much external work could be done by letting the plates come together? Does this equal the energy that was initially stored in the electric field?
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