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A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 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 source so the potential difference between the plates remains constant?

Short Answer

Expert verified
(a) 4.19 J, (b) 16.76 J.

Step by step solution

01

Identify Initial Parameters

The initial energy stored in the capacitor is 8.38 J, and the separation between the plates is 2.30 mm. These are the initial conditions from which we'll calculate the energy in both scenarios.
02

Define Key Equations

The energy stored in a capacitor can be expressed by two equations:1. When charge is constant: \( U = \frac{Q^2}{2C} \) and \( C = \varepsilon \frac{A}{d} \)2. When potential is constant: \( U = \frac{1}{2} C V^2 \) and \( C = \varepsilon \frac{A}{d} \)Here, \( \varepsilon \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.
03

Calculate New Capacitance with Decreased Separation

When the separation is decreased from 2.30 mm to 1.15 mm, the capacitance doubles because capacitance \( C \) is inversely proportional to the separation \( d \). Thus, if initial capacitance is \( C_0 \), the new capacitance \( C_1 = 2C_0 \).
04

Calculate Energy with Constant Charge (Disconnected)

When the capacitor is disconnected, the charge \( Q \) remains constant. With capacitance doubled, the new energy \( U_1 \) becomes\[ U_1 = \frac{Q^2}{2 \times 2C_0} = \frac{1}{2} \cdot \frac{Q^2}{2C_0} = \frac{1}{2} U_0 = \frac{1}{2} \times 8.38 \, \text{J} = 4.19 \, \text{J} \]Thus, the energy becomes 4.19 J.
05

Calculate Energy with Constant Potential (Connected)

When the capacitor remains connected to the potential source, the voltage \( V \) remains constant. With capacitance doubled:\[ U_2 = \frac{1}{2} (2C_0) V^2 = 2 \left(\frac{1}{2} C_0 V^2\right) = 2U_0 = 2 \times 8.38 \, \text{J} = 16.76 \, \text{J} \]Thus, the energy becomes 16.76 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 is a type of capacitor that consists of two parallel conductive plates separated by an insulating material, known as a dielectric. These plates can store electrical energy by holding opposite charges. The key feature of a parallel-plate capacitor is the uniform electric field generated between the plates, which allows it to efficiently store and release energy.
  • Structure: Two conductive plates with a dielectric between them.
  • Function: Stores electrical energy by accumulating charges on the plates.
  • Common application: Used in electronic circuits and devices for energy storage and management.
The capacitance of a parallel-plate capacitor, which indicates how much charge it can store for a given voltage, is determined by the equation \( C = \varepsilon \frac{A}{d} \). Here \( \varepsilon \) is the permittivity of the dielectric material, \( A \) is the area of the plates, and \( d \) is the separation between them. Thus, reducing the distance \( d \) increases the capacitance, meaning the capacitor can store more charge.
Energy storage in capacitors
Capacitors are fascinating devices because they can store electrical energy that can be released quickly when needed. The energy stored in a capacitor is given by different equations depending on the conditions.
  • Constant charge: When a capacitor is disconnected from a voltage source, the charge \( Q \) on the plates remains constant. The energy \( U \) is given by \( U = \frac{Q^2}{2C} \). Decreasing plate separation doubles capacitance, halving the stored energy, as in the equation.
  • Constant potential: If the capacitor is connected to a voltage source, the voltage \( V \) stays constant. The energy is \( U = \frac{1}{2} C V^2 \), so increasing capacitance doubles energy storage.
This dual nature—where energy depends differently on conditions—makes capacitors versatile in different applications like filtering in circuits, smoothing voltage, or providing bursts of energy.
Permittivity of free space
The permittivity of free space, also known as the vacuum permittivity, is a physical constant denoted by \( \varepsilon_0 \). It is crucial in calculating the capacitance of capacitors, especially when the space between the plates is a vacuum or filled with air.
  • Symbol: \( \varepsilon_0 \)
  • Value: Approximately \( 8.854 \times 10^{-12} \, \text{F/m} \) (farads per meter).
  • Role: Determines how much electric field (and thus charge) can be stored between plates in a vacuum.
Inserting this constant into the capacitance formula \( C = \varepsilon \frac{A}{d} \), it helps quantify how effective a capacitor will be at storing energy. Higher permittivity leads to higher capacitance, enabling more charge to be stored for the same physical size. This makes it a fundamental concept in understanding not just capacitors but the broader electromagnetic interactions in vacuum.

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Most popular questions from this chapter

A parallel-plate capacitor has plates with area 0.0225 m\(^2\) separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 V. (b) Use Gauss's law (Eq. 24.23) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

A parallel-plate air capacitor is made by using two plates 12 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm, what are the answers to parts (a)-(d)?

Suppose that the change in V\(_m\) was caused by the entry of Ca\(^{2+}\) instead of Na\(^+\). How many Ca\(^{2+}\) ions would have to enter the cell per unit membrane to produce the change? (a) Half as many as for Na\(^+\); (b) the same as for Na\(^+\); (c) twice as many as for Na\(^+\); (d) cannot say without knowing the inside and outside concentrations of Ca\(^{2+}\).

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

A parallel-plate capacitor has capacitance \(C_0\) = 8.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m? (b) A dielectric with \(K = 2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 \(\times\) 10\(^4\) V/m?

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