Chapter 17: Problem 42
A parallel plate capacitor with air as medium between the plates has a capacitance of \(10 \mu \mathrm{F}\). The area of the capacitor is divided into two equal halves and filled with two media having dielectric constants \(K_{1}=2\) and \(K_{2}=4\). The capacitance of the system will now be (a) \(10 \mu \mathrm{F}\) (b) \(20 \mu \mathrm{F}\) (c) \(30 \mu \mathrm{F}\) (d) \(40 \mu \mathrm{F}\)
Short Answer
Step by step solution
Understanding the Problem
Calculate Capacitor with Dielectric
Divide the Area
Calculate and Sum Capacitance
Solution Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parallel Plate Capacitor
- Structure: Composed of two large, flat plates with a small distance between them.
- Functionality: Stores energy and releases it when needed.
- Factors Affecting Capacitance: The surface area of the plates and the distance between them.
Dielectric Constant
- Effect: It directly influences the capacitance of a capacitor.
- Calculation: The capacitance of a capacitor with a dielectric is the product of its capacitance in a vacuum and the dielectric constant.
- Range: Different materials have different dielectric constants, and this range affects how they are used.
Capacitance Calculation
Process:
- Divide the capacitor area into sections based on the dielectric material used.
- Calculate capacitance for each section: Multiply the original capacitance by the dielectric constant.
- Combine the capacitances of sections in parallel by summing them up.
Electrical Engineering
- Applications: From consumer electronics like smartphones to industrial machinery.
- Importance: Capacitors are fundamental in circuits for filtering, tuning, and energy storage.
- Design Considerations: Engineers often need to balance size, cost, and performance when selecting or designing capacitors.