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24.30. A parallel-plate vacuum capacitor has 8.38 \(\mathrm{J}\) of energy stored in it. The separation between the plates is 2.30 \(\mathrm{mm}\) . If the separation is decreased to 1.15 \(\mathrm{mm}\) , what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential sotrce so the potential difference between the plates remains constant?

Short Answer

Expert verified
For both conditions, the energy stored becomes 16.76 J.

Step by step solution

01

Understanding Given Information

We have a parallel-plate vacuum capacitor with an initial energy of 8.38 J and a plate separation of 2.30 mm. The task is to compute the energy when the separation is reduced to 1.15 mm under two different conditions: charge constant and potential difference constant.
02

Recalling the Energy Equation for Capacitors

The energy stored in a capacitor can be calculated using the equation \( U = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the potential difference. For a parallel-plate capacitor, \( C = \frac{\varepsilon_0 A}{d} \), where \( A \) is the plate area and \( d \) is the separation between the plates.
03

Condition a: Charge Constant

When the capacitor is disconnected, the charge \( Q \) remains constant. The energy equation becomes \( U = \frac{Q^2}{2C} \). With a decreased separation \( d \), the capacitance \( C \) increases (since \( C \propto \frac{1}{d} \)), thus reducing the energy \( U \propto \frac{1}{d} \). Therefore, with \( d_{new} = \frac{1}{2}d_{original} \), the new energy \( U_{new} = 2U_{original} = 2 \times 8.38 = 16.76 \text{ J} \).
04

Condition b: Potential Difference Constant

When connected to the source, the potential difference \( V \) remains constant. The energy \( U \) is proportional to the capacitance \( C \), which increases as \( d \) decreases. Therefore, with \( d = \frac{1}{2}d_{original} \), the capacitance doubles, hence \( U_{new} = 2U_{original} = 2 \times 8.38 = 16.76 \text{ J} \).

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Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor consists of two conductive plates placed parallel to each other, separated by a certain distance. This simple structure is widely used in various electronic devices. How does it work? When a voltage is applied across the plates, an electric field forms, and electric charge accumulates on each plate.
The function of a parallel-plate capacitor can be greatly affected by the separation between the plates. Reducing the distance will increase the capacitor's ability to store charge for a given voltage, because of the stronger electric field. This property allows engineers to design capacitors with desired characteristics for different applications.
Overall, the key features of a parallel-plate capacitor include:
  • Two conductive plates.
  • Electric field formation between plates.
  • Dependence on plate separation for capacitance.
Energy Stored in Capacitors
The energy stored in a capacitor is fundamental to its operation. Capacitors are designed to store and release energy quickly, which is useful in many electronics applications. The energy in a capacitor is given by the formula: \[ U = \frac{1}{2} C V^2 \] where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the potential difference.
In the context of the exercise, when the separation between the plates changes, the energy stored in the capacitor also changes. If the charge remains constant (when disconnected), the increase in capacitance decreases the stored energy proportionally. However, if the voltage remains constant (when connected), the stored energy increases with capacitance.
This demonstrates that energy storage in capacitors is closely related to how the capacitor is integrated into circuits.
  • Depends on capacitance and voltage.
  • Changes with plate separation.
  • Varies with circuit conditions: disconnected vs. connected.
Capacitance Calculations
Calculations of capacitance are crucial for understanding how capacitors function in a circuit. The capacitance, \( C \), of a parallel-plate capacitor can be calculated using: \[ C = \frac{\varepsilon_0 A}{d} \] where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between them.
In the given exercise, decreasing the plate separation leads to an increase in capacitance because \( C \) is inversely proportional to \( d \). Thus, even slight changes in these parameters can have significant effects on the capacitor’s performance.
Understanding and calculating capacitance are key skills for designing circuits with capacitors. Important points include:
  • Directly proportional to plate area.
  • Inversely proportional to plate separation.
  • Influenced by the surrounding medium’s permittivity.
Vacuum Capacitor
Vacuum capacitors are a specialized type of capacitor where the space between the plates is a vacuum, instead of a dielectric material. This makes them excellent for high-voltage applications due to their low loss and high stability characteristics.
In a vacuum capacitor, the vacuum serves as the medium for the electric field, providing several advantages:
  • Minimal energy losses compared to other dielectrics.
  • Greater efficiency in energy storage.
  • Ability to handle high voltages without breakdown.
While they are more expensive and less compact than other types, vacuum capacitors are invaluable in radio frequency and microwave applications where performance is critical.
In the provided exercise, the vacuum plays a role in determining the base capacitance since this particular example assumes no other dielectric is present. Understanding vacuum capacitors helps in recognizing their application in scenarios where high precision and reliability are required.

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Most popular questions from this chapter

24.77. Three square metal plates \(A, B,\) and \(C,\) each 12.0 \(\mathrm{cm}\) on a side and 1.50 \(\mathrm{mm}\) thick, are arranged as in Fig. 24.43 . The plates are separated by sheets of paper 0.45 \(\mathrm{mm}\) thick and with dielectric constant \(4.2 .\) The outer plates are connected together and connected to point \(b\) . The inner plate is connected to point \(a\) (a) Copy the diagram and show by plus and minus signs the charge distribution on the plates when point \(a\) is maintained at a positive potential relative to point \(b\) . (b) What is the capacitance between points \(a\) and \(b ?\)

24.40. A budding electronics hobbyist wants to make a simple 1.0 -nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few shects of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 \(\mathrm{mm}\) . (a) If the sheets of paper measure \(22 \times 28 \mathrm{cm}\) and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of \(120 \mathrm{mm},\) instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 \(\mathrm{nF}\) of capacitance? (c) Suppose she goes hight-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Tellon than of posterboard? Explain.

24.12. A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 \(\mathrm{cm}\) and the capacitance is 116 \(\mathrm{pF}\) . (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 \(\mathrm{V}\) , what is the magnitude of charge on each sphere?

24.6. A \(10.0-\mu F\) parallel-plate capacitor is connected to a \(12.0-\mathrm{V}\) battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if ( i) the plate separation were doubled; (ii) the radius of each plate were doubled and, but their separation was unchanged?

24.13. A spherical capacitor contains a charge of 3.30 \(\mathrm{nC}\) when connected to a potential difference of 220 \(\mathrm{V}\) . If its plates are sepa- rated by vacuum and the inner radius of the outer shell is \(4.00 \mathrm{cm},\) calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

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