Chapter 24: Problem 33
A 12.5-\(\mu\)F capacitor is connected to a power supply that keeps a constant potential difference of 24.0 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?
Short Answer
Step by step solution
Calculate the initial energy stored in the capacitor
Calculate the new capacitance with the dielectric
Calculate the energy stored with the dielectric
Calculate the change in energy
Determine the nature of energy change
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dielectric Constant
In simple terms:
- "Dielectric Constant (\( K \)) = \text{Capacitance with dielectric} / \text{Capacitance without dielectric}"
- The higher the dielectric constant, the greater the increase in capacitance.
- A vacuum has a dielectric constant of 1, while materials like paper or porcelain have higher values due to their ability to store more electrical energy.
Capacitance
Key points to understand capacitance are:
- Measured in Farads (\( F \)), which quantify the charge stored per volt.
- Larger plate areas or reduced distance between plates generally lead to higher capacitance.
- Introducing a dielectric material increases capacitance by reducing the electric field strength between the plates.
Potential Difference
- Measured in Volts (\( V \)). It communicates the pressure from an electric circuit's power source that pushes charged electrons (current) through a conducting loop.
- In capacitors, charge collects on plates of opposite potentials, creating a potential difference.
- In this specific exercise, the potential difference remains constant at 24 V even after the dielectric insertion, showing that energy change occurs due to capacitance alteration, not a difference in voltage.
Energy Stored in Capacitors
- It relies on both the capacitance \( C \) and the potential difference \( V \).
- Adding a dielectric increases the capacitance \( C \), enhancing energy storage \( U \) as the relationship is quadratic with voltage.
- In our scenario, energy increases after dielectric insertion due to increased capacitance, calculated using both pre and post-insertion capacitance values.