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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 air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 cm\(^2\), what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

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.

Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10. (See \(\textbf{Fig. P24.48}\).) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of 85 mV across its membrane. What is the electric field inside this membrane?

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 \(\times\) 10\(^7\) V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00 \(\times\) 10\(^2\) V. The electric field between the plates is to be no greater than 1.00 \(\times\) 10\(^4\) N/C. As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

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