Problem 71 - College Physics

Question

A \(\mathrm{A} 12.5 \mu \mathrm{F}\) capacitor is connected to a power supply that keeps a constant potential difference of 24.0 \(\mathrm{V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

Step-by-Step Solution

Answer

The energy increased by 9.9 mJ after inserting the dielectric.

Step 1: Calculate Initial Energy Without Dielectric

The energy 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 voltage. First, calculate the initial energy with the original capacitance \( C = 12.5 \mu F = 12.5 \times 10^{-6} F \) and voltage \( V = 24.0 V \):\[U_1 = \frac{1}{2} \times 12.5 \times 10^{-6} \times (24.0)^2 = 3.6 \times 10^{-3} \text{ J (or Joules)}\]

Step 2: Determine New Capacitance with Dielectric

When a dielectric is inserted, the new capacitance \( C' \) is given by \( C' = k C \), where \( k \) is the dielectric constant. The dielectric constant given is 3.75. Thus, the new capacitance becomes:\[C' = 3.75 \times 12.5 \times 10^{-6} = 46.875 \times 10^{-6} F\]

Step 3: Calculate Energy with Dielectric

Using the enhanced capacitance \( C' = 46.875 \times 10^{-6} F \) and the constant voltage \( V = 24.0 V \), calculate the energy stored with the dielectric:\[U_2 = \frac{1}{2} \times 46.875 \times 10^{-6} \times (24.0)^2 = 13.5 \times 10^{-3} \text{ J}\]

Step 4: Calculate Energy Change

Calculate the change in energy \( \Delta U \) by subtracting the initial energy from the new energy:\[\Delta U = U_2 - U_1 = 13.5 \times 10^{-3} - 3.6 \times 10^{-3} = 9.9 \times 10^{-3} \text{ J}\]

Step 5: Determine Increase or Decrease

Since the energy increased from \( 3.6 \times 10^{-3} \) J to \( 13.5 \times 10^{-3} \) J, the stored energy in the capacitor increased when the dielectric was inserted.

Key Concepts

Capacitance and dielectrics

When dealing with capacitors, one crucial factor is capacitance, which is a measure of a capacitor's ability to store charge. Capacitance is dependent on the physical characteristics of the capacitor, like the area of the plates and the distance between them. It's typically denoted as \( C \) and measured in Farads. A key concept related to capacitors is the role of dielectrics.
Dielectrics are insulating materials inserted between the plates of a capacitor. They have a significant impact—they increase the capacitor's capacitance.
When a dielectric material is placed between the plates, the new capacitance \( C' \) is calculated using the formula \( C' = kC \), where \( k \) is the dielectric constant. This constant varies with each material, characterizing how effectively it can increase the capacitance of a capacitor.
Dielectrics not only improve capacitance, but they also allow capacitors to operate at higher voltages without risk of breakdown. This ability can be essential in practical applications where capacitors need to store more energy efficiently while maintaining safety standards.

Energy stored in a capacitor

The energy stored in a capacitor is a fundamental concept in electronics and physics. It's the amount of electrical energy that a capacitor can retain, and it depends on the capacitance \( C \) and the voltage \( V \) applied across its plates.
The energy \( U \) stored in a capacitor can be expressed using the formula: \[ U = \frac{1}{2} C V^2 \\] This equation provides a way to calculate energy by taking half of the product of the capacitance and the square of the voltage. This method demonstrates that even a small change in voltage could lead to a significant change in the energy stored.
In our specific example, before the dielectric was inserted, the capacitor had a capacitance of \( 12.5 \, \mu F \) and a voltage of 24.0 \( V \), resulting in an energy of \( 3.6 \times 10^{-3} \text{ J} \). After inserting a dielectric with a constant of 3.75, the capacitance increased, thus increasing the energy stored to \( 13.5 \times 10^{-3} \text{ J} \). This increased the ability of the capacitor to store energy, highlighting the practical usefulness of incorporating dielectrics.

Dielectric constant

The dielectric constant, often denoted as \( k \), is a dimensionless number that describes how much a dielectric material can increase a capacitor's capacitance compared to the capacitance of a vacuum.
This value is pivotal in selecting materials for various electrical applications. A higher dielectric constant means that the material can store more electrical energy and thereby increase the effectiveness of a capacitor.
In our example, the dielectric constant was \( 3.75 \), meaning the material enhanced the capacitance by 3.75 times its original value. This transformation allows capacitors to adapt to more demanding energetic requirements simply by incorporating specific materials.
Overall, understanding the dielectric constant helps engineers and scientists in designing circuits and systems by optimizing the efficiency and effectiveness of capacitors. It's a key factor in material selection, influencing how an electrical device performs in a real-world setting.