Problem 89
Question
Three capacitors have identical geometries. One is filled with a material whose dielectric constant is \(2.50 .\) Another is filled with a material whose dielectric constant is \(4.00 .\) The third capacitor is filled with a material whose dielectric constant \(\kappa\) is such that this single capacitor has the same capacitance as the series combination of the other two. Determine \(\kappa\)
Step-by-Step Solution
Verified Answer
The dielectric constant \( \kappa \) is approximately 1.54.
1Step 1: Understand Capacitance in Series
The capacitance of a series combination of two capacitors is given by the formula: \( \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} \). Since the capacitors have identical geometry and only differ by their dielectric constant, their capacitances can be expressed as proportional to their dielectric constants.
2Step 2: Express Each Capacitance with Dielectric Constants
Let the capacitance of identical geometry without dielectric be \( C_0 \). Thus, for each capacitor: \( C_1 = 2.50 C_0 \) and \( C_2 = 4.00 C_0 \) because the capacitance \( C = \kappa C_0 \).
3Step 3: Calculate Series Capacitance
Using the series formula: \( \frac{1}{C_{series}} = \frac{1}{2.50 C_0} + \frac{1}{4.00 C_0} \). Simplifying, \( \frac{1}{C_{series}} = \frac{0.4 + 0.25}{C_0} = \frac{0.65}{C_0} \). Hence, \( C_{series} = \frac{C_0}{0.65} \).
4Step 4: Determine the Capacitance of the Third Capacitor
The capacitance of the third capacitor filled with material \( \kappa \) is simply \( C_{3} = \kappa C_0 \). Since this equals \( C_{series} \), we have \( \kappa C_0 = \frac{C_0}{0.65} \).
5Step 5: Solve for \( \kappa \)
By equating and simplifying the expressions, we get: \( \kappa = \frac{1}{0.65} \approx 1.54 \). This is the dielectric constant of the third capacitor.
Key Concepts
Dielectric ConstantCapacitance CalculationElectrical Engineering Concepts
Dielectric Constant
The dielectric constant, often symbolized as \(\kappa\), is a crucial part of understanding how capacitors work. It measures a material's ability to increase the capacitance of a capacitor as compared to when a vacuum is used as the dielectric material. In simpler terms, it tells you how much "extra" capacitance a material can provide. For instance, if a material has a dielectric constant of 2.50, it means it can store 2.5 times more electrical energy than a vacuum under the same conditions.
- The dielectric constant is a dimensionless number, meaning it has no units attached to it.
- Materials with higher dielectric constants are often more effective at storing electric charge.
- The greater the dielectric constant, the higher the capacitance of a capacitor for the same geometric structure.
Capacitance Calculation
Capacitance calculation is fundamental for designing and understanding circuits, especially when dealing with capacitors in series. The formula for the total capacitance \( C_{series} \) of capacitors in series is \( \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + ... \). This means that the total capacitance is not simply the sum of the individual capacitances but the reciprocal of the sum of their reciprocals.
- For capacitors with identical geometry, the total capacitance depends primarily on the dielectric constants of the materials used.
- In the given exercise, since each capacitor's capacitance is proportional to its dielectric constant, calculations are simplified.
- Understanding these calculations helps in evaluating how changes in material or configuration will affect the performance of circuits.
Electrical Engineering Concepts
Electrical engineering leverages fundamental concepts about electricity and circuitry, such as the behavior of capacitors in circuits. Capacitors are premium components crucial in storing and managing electrical energy efficiently. A solid grasp of electrical engineering concepts involves knowing how capacitance, resistance, and inductance interact in a circuit.
- In a series circuit, the total capacitance is decreased because the charge has to "jump" from capacitor to capacitor.
- This is crucial in applications where precise control of electrical charge is needed, like in tuning circuits.
- Understanding how to manipulate these concepts enables engineers to design more efficient and effective electronic devices.
Other exercises in this chapter
Problem 86
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