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A three-shell capacitor A capacitor consists of three concentric spherical shells with radii \(R, 2 R\), and \(3 R\). The inner and outer shells are connected by a wire (passing through a hole in the middle shell, without touching it), so they are at the same potential. The shells start neutral, and then a battery transfers charge from the middle shell to the inner and outer shells. (a) If the final charge on the middle shell is \(-Q\), what are the charges on the inner and outer shells? (b) What is the capacitance of the system? (c) If the battery is disconnected, what happens to the three charges on the shells if charge \(q\) is added to the outer shell?

Short Answer

Expert verified
(a) The charge on both the inner and outer shell is \(Q/2\).\n (b) The capacitance of the system can be calculated using the formula for the capacitance of a sphere and the concept of capacitors in series.\n (c) If the battery is disconnected and a charge \(q\) is added to the outer shell, the charges on the three shells will be \(Q/2\), \(-Q - q\) and \(Q/2 + q\) for the inner, middle and outer shell respectively.

Step by step solution

01

Calculate the charges

The inner and outer shells are connected, they have the same potential, and as they were initially neutral and a charge \(-Q\) has been transferred to the middle shell, they must both have a charge of \(Q/2\).
02

Compute the capacitance

Firstly, use the formula for capacitance of a spherical shell: \(C=4 \pi \epsilon (R_2 R_1)/(R_2 - R_1)\) where \(\epsilon\) is the permittivity of free space and \(R_1\) and \(R_2\) are the radii of the shells. We need to compute the total capacitance which can be found by summing up the reciprocals of the capacitances of the outer and inner shells with respect to the middle shell. Thus, the equivalent capacitance becomes \(1/C_{total}=1/C_{out}+1/C_{in}\), where \(C_{out}=4 \pi \epsilon (3R2R)/(3R-2R)\) is the capacitance between the outer and the middle shell and \(C_{in}=4 \pi \epsilon (2RR)/(2R-R)\) is the capacitance between the inner shell and the middle shell.
03

Analyze charge distribution after the battery is disconnected

Once the battery is disconnected, the system of shells is isolated and the total charge is conserved. If you add a charge \(q\) to the outer shell, the charge on the outer shell becomes \(Q/2 + q\). However, the charge on the inner shell does not change because it is isolated from the outer shell; hence it remains \(Q/2\). The total charge on the inner and outer shell becomes \(Q + q\). To maintain electrical neutrality, the middle shell must carry a charge \(-Q - q\).

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Key Concepts

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

Understanding Capacitance in a Three-Shell Capacitor
Capacitance is a fundamental concept when dealing with capacitors. It describes a capacitor's ability to store charge. In our exercise, we have three spherical shells, and their capacitance is an essential parameter to compute.
The capacitance of a simple spherical capacitor is given by the formula \( C = 4 \pi \epsilon \frac{R_2 R_1}{R_2 - R_1} \), where \( \epsilon \) is the permittivity of free space, and \( R_2 \) and \( R_1 \) are the radii of the outer and inner spheres, respectively.
In our problem, the equivalent capacitance is calculated by considering each spherical shell pair's capacitance. Capacitors in combination must be handled with care:
  • First, compute the capacitance between the inner and middle shells using their respective radii.
  • Then, determine the capacitance between the middle and outer shells.
  • Finally, use these individual capacitances to find an equivalent capacitance using the reciprocal formula \( \frac{1}{C_{total}} = \frac{1}{C_{out}} + \frac{1}{C_{in}} \).
This approach ensures we fully understand each shell's role in the overall capacitance of the system.
Exploring Charge Distribution in Spherical Capacitors
Charge distribution in any capacitor system is vital for understanding how it stores and transfers electricity. For our three-shell spherical capacitor, charge is initially balanced, but changes once the battery introduces a charge transfer.
Initially, all shells are neutral. The inner and outer shells are connected, which ensures they remain at the same potential. The battery transfers charge \(-Q\) from the middle shell to both the inner and outer shells.
  • The middle shell loses charge, acquiring \(-Q\).
  • The inner and outer shells share this charge equally due to their connection, each receiving \(Q/2\).
Understanding this distribution helps us track how charges move and establish potential across the system. This configuration maintains the system’s stability and function, given how charges spread.
Maintaining Electrical Neutrality
Electrical neutrality is a principle where the total charge within a closed system remains constant. In our three-shell capacitor, this concept is at the core of what happens during and after charge transfer.
At the beginning, all shells are electrically neutral. Then, the battery assigns a \(-Q\) charge to the middle shell, while the inner and outer shells each get \(Q/2\). When the battery disconnects, the system's total charge is conserved because it is isolated.
If an additional charge \(q\) is added to the outer shell after disconnecting the battery, it becomes \(Q/2 + q\). Though the charge on the inner shell stays the same as \(Q/2\). To maintain neutrality across the system:
  • The total charge within the inner and outer shells becomes \(Q + q\).
  • The middle shell must now have a charge of \(-Q - q\) to compensate for the gain on the outer shell.
This ensures all changes in charge are balanced so the system remains electrically neutral.
Role of Permittivity of Free Space
The permittivity of free space, \( \epsilon \), is a constant that describes how an electric field affects and is affected by a vacuum. This value is crucial for calculations involving electric fields and capacitors.
In capacitors, \( \epsilon \) determines how much charge the structure can hold at a specific voltage. The equation for capacitance highlights \( \epsilon \) prominently, ensuring that the product of \( 4 \pi \epsilon \) directly influences the overall capacitance.
This constant's importance in the spherical capacitor exercise cannot be overstated since it impacts how the electric potential and field strength are handled between the concentric shells. This means:
  • Higher permittivity allows more charge to be stored at a given voltage.
  • Influences how the electric field's lines are distributed across and within the shells.
Understanding \( \epsilon \) helps connect the physics behind electrostatics and how capacitors perform in various settings, making it an invaluable component in our calculations.

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Most popular questions from this chapter

Stationary rod and moving charge A charge \(q\) moves with speed \(v\) parallel to a long rod with linear charge density \(\lambda\), as shown in Fig. 5.30. The rod is at rest. If the charge \(q\) is a distance \(r\) from the rod, the force on it is simply \(F=q E=q \lambda / 2 \pi r \epsilon_{0}\) Now consider the setup in the frame that moves along with the charge \(q\). What is the force on the charge \(q\) in this new frame? Solve this by: (a) transforming the force from the old frame to the new frame, without caring about what causes the force in the new frame; (b) calculating the electric force in the new frame.

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.)

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.

Automobile battery If the voltage at the terminals of an automobile battery drops from \(12.3\) to \(9.8\) volts when a \(0.5\) ohm resistor is connected across the battery, what is the internal resistance?

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