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Magazine Chapter 16: Problem 45
A 50 -pF capacitor is immersed in silicone oil, which has a dielectric constant of \(2.6 .\) When the capacitor is connected to a 24 - \(V\) battery, \((\) a) what will be the charge on the capacitor? (b) How much energy is stored in the capacitor?
Short Answer
Expert verified
(a) 3,120 pC, (b) 37.44 nJ.
Step by step solution
01
Identify the Relevant Equation for Charge
The charge on the capacitor can be calculated using the equation for a capacitor: \[ Q = C \times V \]where \( Q \) is the charge, \( C \) is the capacitance, and \( V \) is the voltage.
02
Adjust Capacitance with Dielectric Constant
When a dielectric material is introduced, the capacitance increases by a factor equal to the dielectric constant. Thus, the new capacitance \( C' \) is given by:\[ C' = \kappa \times C = 2.6 \times 50 \, \text{pF} \]
03
Calculate New Capacitance
Substituting into the equation for the new capacitance:\[ C' = 2.6 \times 50 \, \text{pF} = 130 \, \text{pF} \]
04
Calculate the Charge
Using the updated capacitance value, calculate the charge as:\[ Q = C' \times V = 130 \, \text{pF} \times 24 \, V \ = 3,120 \, \text{pC} \]
05
Identify the Relevant Equation for Energy
The energy stored in a capacitor is given by the equation:\[ U = \frac{1}{2} C' V^2 \]where \( U \) is the energy.
06
Calculate the Energy Stored
Substitute the values into the equation for energy to find:\[ U = \frac{1}{2} \times 130 \, \text{pF} \times (24 \, V)^2 \]Convert 130 pF to farads (130 pF = 130 x 10^-12 F) and calculate:\[ U = \frac{1}{2} \times 130 \times 10^{-12} \, F \times 576 \, V^2 = 37.44 \times 10^{-9} \, J = 37.44 \, \text{nJ} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dielectric Constant
When a material is placed between the plates of a capacitor, it is known as a dielectric. The dielectric constant, often symbolized as \( \kappa \), measures how much a dielectric material can increase the capacitance of a capacitor. Here's how it works:
In our exercise, the silicone oil has a dielectric constant of \(2.6\). This means it increases the capacitance of the original 50 pF capacitor by a factor of \(2.6\). Thus, the new capacitance \( C' \) becomes \( 130 \) pF, effectively illustrating how the dielectric material enhances energy storage capabilities.
- The dielectric constant is a ratio. It compares the permittivity of the dielectric material to the permittivity of a vacuum.
- It's a dimensionless number. In other words, it has no units, making it a simple scalar to multiply with the original capacitance.
- Higher dielectric constants mean better insulating properties. This results in higher capacitance and lower voltage between the plates for a given amount of stored charge.
In our exercise, the silicone oil has a dielectric constant of \(2.6\). This means it increases the capacitance of the original 50 pF capacitor by a factor of \(2.6\). Thus, the new capacitance \( C' \) becomes \( 130 \) pF, effectively illustrating how the dielectric material enhances energy storage capabilities.
Energy Stored in Capacitor
The energy stored in a capacitor is an important concept, especially in electrical circuits, as it determines how much energy the capacitor can release when needed. The formula to calculate this is:
In our example, we use the updated capacitance value due to the silicone oil, \( C' = 130 \, \text{pF} \), and a voltage \( V = 24 \, \text{V} \). Substituting these into the formula gives:
- \( U = \frac{1}{2} C' V^2 \)
- \( C' \) is the capacitance with the dielectric.
- \( V \) is the voltage across the capacitor.
In our example, we use the updated capacitance value due to the silicone oil, \( C' = 130 \, \text{pF} \), and a voltage \( V = 24 \, \text{V} \). Substituting these into the formula gives:
- Convert the capacitance to farads: \( 130 \, \text{pF} = 130 \times 10^{-12} \, \text{F} \).
- Calculate the energy stored: \[U = \frac{1}{2} \times 130 \times 10^{-12} \, \text{F} \times (24 \, \text{V})^2 = 37.44 \times 10^{-9} \, \text{J} = 37.44 \, \text{nJ}\]
Capacitance
Capacitance is the ability of a system to store an electric charge. It is a fundamental concept in learning about capacitors. Capacitance is expressed by the formula:
In real-world terms, if you increase the surface area of the plates or decrease the distance between them, you can enhance the capacitance. Furthermore, by adding a dielectric material:
In the exercise, the initial capacitance of the capacitor was 50 pF. However, when the silicone oil (with its dielectric constant of 2.6) is added, the effective capacitance is increased to 130 pF. This demonstrates how materials and design can influence the efficiency and capacity of capacitors in practical applications.
- \( C = \frac{Q}{V} \)
- \( C \) is the capacitance in farads (F), \( Q \) is the charge in coulombs, and \( V \) is the voltage in volts.
In real-world terms, if you increase the surface area of the plates or decrease the distance between them, you can enhance the capacitance. Furthermore, by adding a dielectric material:
- **Capacitance Increases:** The additive effect of dielectrics is clear as they increase a capacitor's capacitance by reducing the electric field (thus allowing it to store more charge).
- **Energy Efficiency:** Higher capacitance means more energy storage for the same voltage, which is beneficial for electronic devices.
In the exercise, the initial capacitance of the capacitor was 50 pF. However, when the silicone oil (with its dielectric constant of 2.6) is added, the effective capacitance is increased to 130 pF. This demonstrates how materials and design can influence the efficiency and capacity of capacitors in practical applications.
