Question

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

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

Step-by-step solution

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.

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.

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 \( C_1 = 2C_0 \).

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.

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.

Key concepts

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.