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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

Expert verified
The new capacitance is 30 µF (option c).

Step by step solution

01

Understanding the Problem

We have a parallel plate capacitor initially with air as the medium, having a capacitance of 10 µF. The area is then divided into two equal parts with two different dielectrics having dielectric constants \( K_1 = 2 \) and \( K_2 = 4 \). We need to find the new capacitance of the capacitor when filled with these dielectrics.
02

Calculate Capacitor with Dielectric

A capacitor with a dielectric increases its capacitance by the dielectric constant. For each segment, the capacitance is multiplied by the dielectric constant of that segment.
03

Divide the Area

Divide the capacitor area into two equal halves. Since the dielectric constant affects capacitance, we need to express the new capacitance configurations. After adding the dielectrics, each half acts as a capacitor in parallel with others, thus we use the formula \( C = \frac{C_0}{2} \cdot K_i \) for each segment.
04

Calculate and Sum Capacitance

Calculate the capacitance of each part which is \( C_1 = \frac{C_0}{2} \, K_1 \) and \( C_2 = \frac{C_0}{2} \, K_2 \). Then, \( C_1 = 5 \, \mu F \times 2 = 10 \, \mu F \) and \( C_2 = 5 \, \mu F \times 4 = 20 \, \mu F \). Sum these capacitances: \( C = C_1 + C_2 = 10 \, \mu F + 20 \, \mu F = 30 \, \mu F \).
05

Solution Conclusion

Since the capacitor segments are in parallel, we add their capacitances. The new capacitance of the system after adding both dielectrics is 30 µF.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel Plate Capacitor
Parallel plate capacitors are one of the simplest types of capacitors used in electrical engineering. They consist of two metal plates, parallel to each other, separated by a medium called a dielectric. The capacitor stores electrical energy in the form of an electric field between the two plates. The ability of a parallel plate capacitor to store charge is quantified by its capacitance, which is measured in farads.
  • 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.
When a voltage is applied across the plates, an electric field is created, storing energy. This configuration allows for efficient energy storage in electronic circuits.
Dielectric Constant
The dielectric constant, also known as the relative permittivity, is a measure of a material's ability to concentrate electric flux. In other words, it indicates how effectively a material can store electrical energy in an electric field. Higher dielectric constants mean the material can store more electric charge at a given voltage.
  • 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.
For the problem in question, two dielectrics with constants 2 and 4 are used, which influences the overall capacitance of the system.
Capacitance Calculation
Calculating capacitance in systems with dielectrics is an important task in electrical engineering. When a dielectric is introduced, the capacitance of a system changes, due to the dielectric constant of the material. For a parallel plate capacitor divided into areas with different dielectrics, we calculate the capacitance for each part separately and then sum them.

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.
In our current exercise, the two halves of the capacitor have dielectrics of constants 2 and 4 leading to capacitances of 10 µF and 20 µF respectively, giving a total new capacitance of 30 µF.
Electrical Engineering
Electrical engineering involves the study and application of electricity, electronics, and electromagnetism. It plays a crucial role in designing and building devices and systems that use electric power. Understanding capacitors and how they function is essential in this field, especially for energy storage solutions.
  • 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.
The capability to manipulate capacitance through materials and design is foundational to advancing electrical engineering solutions, whether for enhancing battery life or developing new digital technologies.

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Most popular questions from this chapter

Two parallel metal plates carry charges \(+Q\) and \(-Q\) respectively. A test charge \(q_{o}\) placed between them experiences a force \(F\). Now the separation between the plates is doubled, then the force on the test charge will be (a) \(\bar{F}\) (b) \(2 F\) (c) \(F / 2\) (d) \(F / 4\)

Two identical air-filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch \(S\). If now the switch \(S\) is opened and the space between the plates filled with a dielectric of specific inductive capacity or relative permittivity 3, then (a) the potential difference as well as charge on each capacitor goes down by a factor of 3 (b) the potential difference as well as charge on each capacitor goes up by a factor of 3 (c) the potential difference across \(B\) remains constant and charge on \(A\) remains unchanged (d) the potential difference across \(A\) remains constant and charge on \(B\) remains unchanged

Suppose \(n\) identical capacitors are joined in parallel and charged to potential \(V .\) Now, they are separated and joined in series. If the energy possessed by each capacitor is \(U\), then on joining them in series, the energy and potential difference for the combination are (a) \(n U, V\) (b) \(U, n V\) (c) \(n U, n V\) (d) less than \(n U, n V\)

Two capacitors, \(3 \mu \mathrm{F}\) and \(4 \mu \mathrm{F}\), are individually charged across a 6 volt battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the common potential? (a) 6 volt (b) \((6 / 7)\) volt (c) 2 volt (d) none of these

1000 small water drops each of radius \(r\) and charge \(q\) coalesce together to form one spherical drop. The potential of the bigger drop is larger than that of the smaller one by a factor (a) 1000 (b) 100 (c) 10 (d) 1

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