Problem 61 - College Physics

Question

A parallel-plate vacuum capacitor has 8.38 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 source so the potential difference between the plates remains constant?

Step-by-Step Solution

Answer

Under constant charge, the energy is 4.19 J; under constant voltage, it's 16.76 J.

Step 1 - Identify Key Variables

We are given an initial energy of 8.38 J stored in a parallel-plate capacitor with an initial plate separation of 2.30 mm. We need to find the energy stored when the separation is reduced to 1.15 mm under two different conditions - constant charge and constant voltage.

Step 2 - Constant Charge: Calculate New Energy

When the capacitor is disconnected, the charge remains constant. We know that energy in a capacitor is given by \( U = \frac{1}{2} \frac{Q^2}{C} \), and capacitance \( C \) for a parallel capacitor is \( \frac{\varepsilon A}{d} \), where \( \varepsilon \) is the permittivity, \( A \) is the area, and \( d \) is the distance between plates. As \( d \) becomes \( \frac{d}{2} \), capacitance doubles, making energy \( U \) half according to the formula. Hence, the new energy will be \( \frac{8.38}{2} = 4.19 \text{ J} \).

Step 3 - Constant Voltage: Calculate New Energy

If the capacitor remains connected to the power source, the voltage is constant. In this case, energy \( U \) is given by \( U = \frac{1}{2} CV^2 \). Since capacitance doubles when separation is halved, the energy stored also doubles to \( 2 \times 8.38 = 16.76 \text{ J} \).

Step 4 - Result Verification

For constant charge conditions, reducing the plate separation decreases the energy to 4.19 J. Conversely, under constant voltage conditions, energy increases to 16.76 J. This confirms changes in energy due to varying capacitance conditions.

Key Concepts

Parallel-plate capacitor

A parallel-plate capacitor consists of two large, flat conductive plates that are parallel to each other with a small gap or space between them. This configuration is designed to store electrical energy in an electric field created between the plates. The plates accumulate charges of equal magnitude but opposite signs, creating an electric field between them. Factors affecting the capacitor's behavior include:

• A: The area of each plate
• d: The distance between the plates
• \(\varepsilon\): The permittivity of the material between the plates

These factors influence the capacity of the capacitor to store charges, and changes to these parameters alter the voltage and energy stored.

Energy storage

Capacitors store energy in the form of an electric field. The amount of energy stored in a parallel-plate capacitor can be expressed by the formula: \[ U = \frac{1}{2} CV^2 \] where \( U \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage across the plates. Another useful expression is: \[ U = \frac{1}{2} \frac{Q^2}{C} \] where \(Q\) is the charge.When designing capacitors or analyzing their behavior, understanding how they store energy helps in knowing how they'll react under changing conditions like altering the distance between plates.

Charge and voltage

Charge and voltage are essential elements of capacitors. Charge (\(Q\)) is the amount of electric charge stored on each plate. Voltage (\(V\)) is the electric potential difference between the plates. With constant charge, (like when disconnected from a power source), the energy within the capacitor depends inversely on capacitance, as seen in the formula: \[ U = \frac{1}{2} \frac{Q^2}{C} \] For constant voltage conditions, such as being connected to a power source, the energy depends directly on capacitance:\[ U = \frac{1}{2} CV^2 \] Understanding the relationship between charge, voltage, and energy helps in predicting how energy changes when adjusting the capacitor's physical parameters.

Permittivity

Permittivity \((\varepsilon)\) is a measure of how easily a material allows the formation of an electric field within it. In capacitors, permittivity influences the capacitance. Higher permittivity materials allow for greater charge storage capacity.In a vacuum, the permittivity \((\varepsilon_0)\) is a constant constructed by the vacuum of free space, approximately equal to \(8.85 \times 10^{-12} \ \, F/m\). Inserting a material with greater permittivity than a vacuum increases capacitance.Mathematically, it's part of the capacitance formula: \[ C = \frac{\varepsilon A}{d} \] Higher permittivity increases capacitance meaning more charge can be stored relative to a given voltage.

Capacitance calculation

Capacitance (\(C\)) is the ability of a capacitor to store a charge per unit voltage. For a parallel-plate capacitor, its value is calculated using the formula:\[ C = \frac{\varepsilon A}{d} \]where:

• \(\varepsilon\) is the permittivity of the dielectric material between the plates
• \(A\) is the area of the plates
• \(d\) is the distance between the plates

Increasing plate area \((A)\) or permittivity \((\varepsilon)\) will increase capacitance, meaning the capacitor can hold more charge. Conversely, increasing distance \((d)\) decreases capacitance. These relationships are crucial for designing capacitors for specific energy storage capabilities and understanding their performance in electronic circuits.