Problem 35 - University Physics with Modern Physics

Question

A 12.5 -\muF 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

(a) Energy before: 3.6 mJ; Energy after: 13.5 mJ. (b) Energy increased by 9.9 mJ.

Step 1: Calculate Initial Energy without Dielectric

The energy stored in a capacitor without a dielectric is given by the formula \( U = \frac{1}{2} C V^2 \). Here, \( C = 12.5 \text{\,}\mu\text{F} = 12.5 \times 10^{-6} \text{\,F} \) and \( V = 24.0 \text{\,V} \). Calculate the initial energy \( U_i \): \[ U_i = \frac{1}{2} \times 12.5 \times 10^{-6} \times (24.0)^2 = 3.6 \times 10^{-3} \text{\,J} \]

Step 2: Compute Capacitance with Dielectric

The capacitance with a dielectric is given by \( C' = \kappa C \), where \( \kappa = 3.75 \) is the dielectric constant. Therefore, the new capacitance \( C' \) is: \[ C' = 3.75 \times 12.5 \times 10^{-6} = 46.875 \times 10^{-6} \text{\,F} \]

Step 3: Calculate Energy with Dielectric Inserted

Use the energy formula \( U' = \frac{1}{2} C' V^2 \) to find the energy stored in the capacitor with the dielectric: \[ U' = \frac{1}{2} \times 46.875 \times 10^{-6} \times (24.0)^2 = 13.5 \times 10^{-3} \text{\,J} \]

Step 4: Determine Energy Change and Direction

The change in energy due to the insertion of the dielectric is given by \( \Delta U = U' - U_i \). Calculate \( \Delta U \) to see if the energy increased or decreased: \[ \Delta U = 13.5 \times 10^{-3} \text{J} - 3.6 \times 10^{-3} \text{J} = 9.9 \times 10^{-3} \text{J} \] Since \( \Delta U \) is positive, the energy increased.

Key Concepts

Dielectric Constant

The dielectric constant, often symbolized by \( \kappa \), is an important concept in understanding capacitors. It is a measure of a material's ability to increase the capacitance of a capacitor when placed between its plates. The dielectric constant is a dimensionless quantity, indicating how many times the capacitance can increase due to the material.

For instance, in our exercise, the dielectric constant is 3.75. This means that when this material is placed between the capacitor plates, it increases the capacitor's capacitance 3.75 times compared to when there is just air or vacuum. The presence of a dielectric also affects the electric field, reducing it within the material and thus allowing the capacitor to store more charge at the same voltage.

This concept is utilized in many applications like electronic circuits and can play a crucial role in enhancing the efficiency and effectiveness of capacitors in various devices.

Energy Stored in a Capacitor

The energy stored in a capacitor relates directly to its ability to hold charge under a certain voltage. This energy can be calculated using the formula \( U = \frac{1}{2} C V^2 \), which helps determine how much electrical energy can be stored before the dielectric is added.

Initially, without a dielectric, the capacitor had a capacitance of 12.5 \( \mu \text{F} \), and a potential difference or voltage \( V \) of 24 volts is maintained across it. Using the formula, the energy stored was computed to be 3.6 milli-joules (\( \text{mJ} \)).

Once the dielectric is introduced, its ability to store energy changes because the capacitance increases. With the new capacitance calculated as 46.875 \( \mu \text{F} \) (after the dielectric constant factor), the energy stored increased to 13.5 \( \text{mJ} \). This demonstrates how dielectric materials effectively enhance a capacitor's storing capability.

Change in Energy

Observing how energy changes when a dielectric is added to a capacitor is crucial to understanding the underlying physics. The difference in energy, denoted as \( \Delta U \), gives insight into how much more energy is stored or released when the dielectric material is introduced.

In this exercise, \( \Delta U = U' - U_i \), where \( U' \) is the energy with the dielectric, and \( U_i \) is the energy without it. The calculation showed an increase in energy from 3.6 \( \text{mJ} \) to 13.5 \( \text{mJ} \), leading to a \( \Delta U \) of 9.9 millijoules (\( \text{mJ} \)).

This increase signifies that more energy can be stored when a dielectric is added. The dielectric stabilizes the charges within the capacitor by reducing the amount of energy lost through the electric field. This has practical implications in various technologies where efficient energy storage is essential. Understanding these changes helps in designing better capacitive systems for electronic devices.