Problem 43 - University Physics with Modern Physics

Question

When a 360 -nF air capacitor \(\left(1 \mathrm{nF}=10^{-9} \mathrm{F}\right)\) is connected to a power supply, the energy stored in the capacitor is \(1.85 \times 10^{-5} \mathrm{J}\) . While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by \(2.32 \times 10^{-5} \mathrm{J}\) . (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

Step-by-Step Solution

Answer

(a) Calculate \(V\) using the initial energy. (b) Use increased energy to find \(C_{\text{new}}\), then \(\kappa = \frac{C_{\text{new}}}{C}\).

Step 1: Understand the Problem

We are given a capacitor with a capacitance of 360 nF connected to a power supply, storing an initial energy of \(1.85 \times 10^{-5}\, \text{J}\). A dielectric increases the stored energy by \(2.32 \times 10^{-5}\, \text{J}\). We need to find the potential difference across the plates and the dielectric constant.

Step 2: Recall the Energy Formula for a Capacitor

The energy \(U\) stored in a capacitor is given by the formula: \[U = \frac{1}{2} C V^2\] where \(C\) is the capacitance and \(V\) is the potential difference.

Step 3: Calculate the Initial Potential Difference

Substituting the initial values into the energy formula, we have: \[ 1.85 \times 10^{-5} = \frac{1}{2} \times 360 \times 10^{-9} \times V^2\]Solve for \(V^2\) and then \(V\): \[ V^2 = \frac{2 \times 1.85 \times 10^{-5}}{360 \times 10^{-9}}\]\[ V = \sqrt{\frac{2 \times 1.85 \times 10^{-5}}{360 \times 10^{-9}}} \]Calculate \(V\) to find the potential difference.

Step 4: Compute Additional Energy Stored with Dielectric

The total energy stored with the dielectric now is the sum of the initial and additional stored energies: \[ U_{\text{total}} = 1.85 \times 10^{-5} + 2.32 \times 10^{-5} = 4.17 \times 10^{-5} \, \text{J}\]

Step 5: Determine the New Capacitance

With the dielectric, the energy formula remains: \[ U_{\text{total}} = \frac{1}{2} C_{\text{new}} V^2\]Solving for \(C_{\text{new}}\):\[ C_{\text{new}} = \frac{2 \times 4.17 \times 10^{-5}}{V^2} \]Use the previously computed \(V\) to determine \(C_{\text{new}}.\)

Step 6: Calculate Dielectric Constant

The dielectric constant \(\kappa\) is defined as the ratio of the new capacitance to the initial capacitance: \[ \kappa = \frac{C_{\text{new}}}{C} \]Insert values to find \(\kappa\).

Key Concepts

Capacitance

Capacitance is a fundamental concept in the study of capacitors. It measures a capacitor's ability to store electrical charge. Essentially, it tells us how much charge a capacitor can hold for a given potential difference across its plates. The formula for capacitance is:

• The formula: \( C = \frac{Q}{V} \)

where \( C \) is the capacitance in farads, \( Q \) is the electric charge in coulombs, and \( V \) is the potential difference in volts.
In this specific scenario, the capacitance of the air-filled capacitor is given as 360 nF (nanofarads). Nanofarads are commonly used in problems involving capacitors in circuits because they represent a manageable size for many electrical components. Capacitors play a crucial role in storing energy temporarily. Evaluating and understanding their capacitance helps determine how effective they are in different applications.
When discussing capacitance, it is crucial to understand how changes to the dielectric (the insulating material situated between the plates) affect it. A dielectric material can increase the capacitance without changing the size or spacing of the plates.

Potential Difference

The potential difference (also known as voltage) is the measure of the energy difference per charge unit between two points in an electric circuit. In a capacitor, it indicates the electrical pressure that pushes the charge into the plates.
To understand the significance of the potential difference, consider the formula used to determine the energy stored in a capacitor:

• \( U = \frac{1}{2} C V^2 \)

where \( U \) is the energy stored, \( C \) is the capacitance, and \( V \) is the potential difference.
In the exercise, the initial potential difference can be deduced by rearranging the equation to solve for \( V \). By substituting known values of energy \( (1.85 \times 10^{-5} \text{ J}) \) and capacitance \( (360 \times 10^{-9} \text{ F}) \), the potential difference that initially existed across the capacitor's plates can be calculated. This calculation is crucial as it serves as the baseline for determining how the dielectric changes the system.

Capacitors and Dielectrics

Capacitors are devices capable of storing electrical energy by means of an electric field present between a set of conductive plates. When a dielectric material is introduced between these plates, it significantly alters the behavior and efficiency of the capacitor.
The dielectric constant, \( \kappa \), is a critical value that measures a material's ability to increase the capacitance relative to a vacuum. Dielectrics enhance a capacitor's capability to store charge because they reduce the electric field strength and increase capacitance. The dielectric constant is evaluated using the relation:

• \( \kappa = \frac{C_{\text{new}}}{C} \)

Given that the dielectric increases the capacitance without changing the physical structure of the capacitor, it demonstrates the effectiveness of dielectrics in practical applications.
Inserting a dielectric like in the exercise provided boosts stored energy from \(1.85 \times 10^{-5} \text{ J} \) to \(4.17 \times 10^{-5} \text{ J} \), offering practical insights into their usage in electronic devices. This increase in energy without altering the power supply showcases why dielectrics are valuable for efficient capacitor function.