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Chapter 26: Problem 58

A 2.00 -nF parallel-plate capacitor is charged to an initial potential difference \(\Delta V_{i}=100 \mathrm{V}\) and then isolated. The dielectric material between the plates is mica, with a dielectric constant of 5.00 . (a) How much work is required to withdraw the mica sheet? (b) What is the potential difference of the capacitor after the mica is withdrawn?

Short Answer

Expert verified
The work required to withdraw the mica sheet is \( W_{i} - W_{f} \) joules, and the final potential difference after the mica is withdrawn is 500 V.

Step by step solution

01

Calculate Initial Energy Stored

Calculate the initial energy stored in the capacitor with the dielectric using the formula, \( W_{i} = \frac{1}{2} C_{i} \Delta V_{i}^2 \) where \( C_{i} \) is the initial capacitance and \( \Delta V_{i} \) is the initial potential difference. The initial capacitance with the dielectric is given by \( C_{i} = k \cdot C_{0} \) where \( k \) is the dielectric constant and \( C_{0} \) is the capacitance without the dielectric. Here, \( C_{0} = 2.00 \) nF and \( k = 5.00 \) so \( C_{i} = 5.00 \cdot 2.00 \) nF = 10.00 nF. Thus, \( W_{i} = \frac{1}{2} \cdot 10.00 \cdot 10^{-9} \cdot 100^2 \) joules.
02

Calculate Final Energy Stored

Calculate the final energy stored in the capacitor without the dielectric using the formula, \( W_f = \frac{1}{2} C_{0} \Delta V_f^2 \) where \( \Delta V_f \) is the final potential difference. Since the capacitor is isolated, the charge stays constant, and \( Q = C_{i} \Delta V_{i} = C_{0} \Delta V_f \) so \( \Delta V_f = \frac{C_{i}}{C_{0}} \Delta V_{i} \) which equals \( \frac{10.00}{2.00} \cdot 100 \) V = 500 V. Then calculate \( W_{f} = \frac{1}{2} \cdot 2.00 \cdot 10^{-9} \cdot 500^2 \) joules.
03

Calculate Work Done to Withdraw the Mica

The work required to withdraw the mica is the difference between the initial and final energies of the capacitor. So, \( W = W_{i} - W_{f} \). From Steps 1 and 2, compute the work done to withdraw the mica, which represents the decrease in energy as the mica is removed.
04

Calculate Final Potential Difference

The final potential difference \( \Delta V_{f} \) has been determined in Step 2 while ensuring that the charge remains constant during mica removal. It can now be reported as the final answer to part (b) of the question.

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Key Concepts

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

Dielectric Constant
The dielectric constant, often symbolized as 'k', measures how much a certain material can reduce the electric field compared to the vacuum. This material is known as a dielectric. Essentially, a dielectric material inserted between the plates of a capacitor increases its capacitance, effectively allowing the capacitor to store more charge for the same potential difference or voltage.

As in the original exercise, mica was used as the dielectric with a constant of 5.00. This number signifies that mica can increase the capacitance of the capacitor by a factor of 5 compared to a vacuum. Capacitance with a dielectric, denoted as \( C_{i} \), is calculated by the formula \( C_{i} = k \times C_{0} \), where \( C_{0} \) is the capacitance without the dielectric. This ability to hold more charge is crucial, as it impacts the amount of energy the capacitor can store and changes the work needed to remove the dielectric, as seen in the related exercise.
Energy Stored in a Capacitor
A capacitor stores electrical energy when it holds a charge separated across its plates. The energy stored in a capacitor can be described by the equation \( W = \frac{1}{2} C \times (\text{potential difference})^2 \). When the potential difference, or voltage across the capacitor's plates, is known, and its capacitance is given, the energy stored can be computed.

In our exercise, the initial energy stored in the capacitor, \( W_{i} \), takes into account the increased capacitance due to the presence of the dielectric. After the mica is removed, the final energy stored is less due to the reduced capacitance. The work required to remove the mica from the capacitor corresponds to the change in the energy stored, which gives us a tangible understanding of how dielectrics influence energy storage in capacitors.
Potential Difference
Potential difference, often referred to as voltage, is the measure of the electrical potential energy difference per unit charge between two points in an electric circuit. The greater the potential difference, the more potential energy per charge an electric charge has when moving between these two points.

In our scenario, the initial potential difference, \( \text{螖}V_{i} \), is provided and is necessary to determine the initial and final energy stored in the capacitor. Due to the law of conservation of charge, when the dielectric is removed and the capacitor is isolated (meaning no charge is allowed to enter or leave the system), the potential difference must change accordingly to ensure the charge remains constant. The formula \( \text{螖}V_{f} = \frac{C_{i}}{C_{0}} \text{螖}V_{i} \) expresses this relationship. This change in potential difference is seminal in understanding electronic behavior in isolated systems, such as capacitors being manipulated in circuits.

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Most popular questions from this chapter

The general form of Gauss's law describes how a charge creates an electric field in a material, as well as in vacuum. It is $$\oint \mathbf{E} \cdot d \mathbf{A}=\frac{q}{\epsilon}$$ where \(\epsilon=\kappa \epsilon_{0}\) is the permittivity of the material. (a) A sheet with charge \(Q\) uniformly distributed over its area \(A\) is surrounded by a dielectric. Show that the sheet creates a uniform electric field at nearby points, with magnitude \(E=Q / 2 A \epsilon\) . (b) Two large sheets of area \(A,\) carrying opposite charges of equal magnitude \(Q,\) are a small distance \(d\) apart. Show that they create uniform electric field in the space between them, with magnitude \(E=Q / A \epsilon\) (c) Assume that the negative plate is at zero potential. Show that the positive plate is at potential \(Q d / A \epsilon\) (d) Show that the capacitance of the pair of plates is \(A \epsilon / d=\kappa A \epsilon_{0} / d .\)

A wafer of titanium dioxide \((\kappa=173)\) of area 1.00 \(\mathrm{cm}^{2}\) has a thickness of 0.100 \(\mathrm{mm}\) . Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a \(12.0-\mathrm{V}\) battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part \((\mathrm{b}),\) what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?

A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How many capacitors are in the group?

A parallel-plate capacitor of plate separation \(d\) is charged to a potential difference \(\Delta V_{0} .\) A dielectric slab of thickness \(d\) and diectric constant \(\kappa\) is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is \(U / U_{0}=\kappa\) Give a physical explanation for this increase in stored energy. (b) What happens to the charge on the capacitor? (Note that this situation is not the same as in Example 26.7 , in which the battery was removed from the circuit before the dielectric was introduced.)

According to its design specification, the timer circuit delaying the closing of an elevator door is to have a capacitance of 32.0\(\mu \mathrm{F}\) between two points \(A\) and \(B .\) (a) When one circuit is being constructed, the inexpensive but durable capacitor installed between these two points is found to have capacitance 34.8\(\mu \mathrm{F}\) . To meet the specification, one additional capacitor can be placed between the two points. Should it be in series or in parallel with the \(34.8-\mu \mathrm{F}\) capacitor? What should be its capacitance? (b) What If? The next circuit comes down the assembly line with capacitance 29.8\(\mu \mathrm{F}\) between \(A\) and \(B\) . What additional capacitor should be installed in series or in parallel in that circuit, to meet the specification?
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