/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 The capacitance of a parallel pl... [FREE SOLUTION] | 91Ó°ÊÓ

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The capacitance of a parallel plate condenser does not depend upon (a) area of the plates (b) medium between the plates (c) distance between the plates (d) metal of the plates

Short Answer

Expert verified
The capacitance does not depend on the metal of the plates.

Step by step solution

01

Introduction to Capacitance

The capacitance of a parallel plate condenser is the ability of the condenser to store electrical charge. It is given by the formula: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_r \) is the relative permittivity of the medium between the plates, \( A \) is the area of one of the plates, and \( d \) is the distance between the plates.
02

Identifying Relevant Factors

From the formula \( C = \frac{\varepsilon_0 \varepsilon_r A}{d} \), we can identify the factors affecting the capacitance:1. \( A \) - Area of the plates: Capacitance is directly proportional to the area of the plates.2. \( \varepsilon_r \) - Medium between the plates: Capacitance depends on the relative permittivity of the medium.3. \( d \) - Distance between the plates: Capacitance is inversely proportional to the distance between the plates.These are factors that directly affect the capacitance value.
03

Understanding the Impact of Metal Type

The metal of the plates is not present in the capacitance formula, \( C = \frac{\varepsilon_0 \varepsilon_r A}{d} \). This indicates that the type of metal used to make the plates does not directly affect the capacitance. The function of the metal is primarily to conduct charge and is assumed to be ideal without impacting stored capacitance due to its conductive properties.
04

Conclusion

From the analysis above, we conclude that the capacitance of a parallel plate condenser does not depend upon the metal of the plates. Therefore, the correct answer to this exercise is (d) metal of the plates.

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Key Concepts

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

Parallel Plate Capacitor
A parallel plate capacitor is a fundamental component in electronics that allows for the storage of electrical charge. It consists of two parallel conductive plates separated by a certain distance with an insulating material, known as a dielectric, between them.

The capacitance of this type of capacitor is determined by the formula: \[ C = \frac{\varepsilon_0 \varepsilon_r A}{d} \]where:
  • \( C \) is the capacitance in farads.
  • \( \varepsilon_0 \) is the permittivity of free space, a constant.
  • \( \varepsilon_r \) is the relative permittivity of the dielectric between the plates.
  • \( A \) is the area of one of the plates in square meters.
  • \( d \) is the distance between the plates in meters.


The key design of a parallel plate capacitor allows it to store energy efficiently. When a voltage is applied across the plates, one plate accumulates positive charge, while the other accumulates an equal amount of negative charge. The energy is stored in the electric field between these plates.
Relative Permittivity
Relative permittivity, also known as the dielectric constant, describes how well a medium can support an electric field within it compared to the electrical vacuum. It is a dimensionless number and directly affects the capacitance of a capacitor.

In the context of the parallel plate capacitor formula \( C = \frac{\varepsilon_0 \varepsilon_r A}{d} \), the relative permittivity \( \varepsilon_r \) multiplies the ability of the capacitor to store charge for a given electric field.

Materials with high relative permittivity help increase capacitance, enhancing the capacitor's ability to store more charge at a lower voltage. In practical applications, different materials are selected based on their dielectric properties to achieve the desired performance in circuit applications.

To explore:
  • Air has a relative permittivity slightly greater than 1.
  • Materials like ceramic or certain polymers may have much higher values, making them suitable for high-capacitance applications.
Electric Field
An electric field is a region around a charged object where other charged objects experience a force. In the case of a parallel plate capacitor, the electric field plays a crucial role in defining its operation.

When a potential difference or voltage is applied across the plates, an electric field is established in the space between the plates. The strength of this electric field \( E \) between the plates is given by:\[ E = \frac{V}{d} \]where:
  • \( V \) is the voltage applied across the capacitor.
  • \( d \) is the distance between the plates.


The electric field is directly related to the voltage and inversely proportional to the separation distance of the plates. It indicates how much force a charge would experience per unit charge placed within the field.

The energy stored within a capacitor is housed in this electric field, and the uniform distribution of the field between the plates is essential for stable capacitor performance. Understanding the electric field helps in designing capacitors that operate efficiently under various electrical conditions.

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Most popular questions from this chapter

Two spherical conductors \(A\) and \(B\) of radii \(a\) and \(b\) \((b>a)\) are placed concentrically in air. \(A\) is given a charge \(+Q\) while \(B\) is earthed. Then the equivalent capacitance of the system is (a) \(4 \pi \varepsilon_{o}\left(\frac{a b}{b-a}\right)\) (b) \(4 \pi \varepsilon_{e}(a+b)\) (c) \(4 \pi \varepsilon_{o} b\) (d) \(4 \pi \varepsilon_{o}\left(\frac{b^{2}}{b-a}\right)\)

The capacitance of a parallel plate capacitor is \(2 \mu \mathrm{F}\) and the charge on its positive plate is \(2 \mu \mathrm{C}\). If the charges on its plates are doubled, the capacitance of the capacitor (a) becomes \(4 \mu \mathrm{F}\) (b) becomes \(1 \mu \mathrm{F}\) (c) remains \(2 \mu \mathrm{F}\) (d) cannot be ascertained in the absence of information on potential

Two identical capacitors are joined in parallel, charged to a potential \(V\), separated and then, connected in series, i.e., the positive plate of one is connected to the negative plate of the other. Then (a) the charges on the free plates connected together are destroyed (b) charges on the free plates are destroyed (c) the energy stored in the system increases (d) the potential difference between the free plates is \(2 \mathrm{~V}\)

A capacitor of capacitance \(C_{1}\) is charged to a potential \(V_{o} .\) The electrostatic energy stored in it is \(U_{o} .\) It is connected to another uncharged capacitor of capacitance \(C_{2}\) in parallel. The energy dissipated in the process is (a) \(\frac{C_{2}}{C_{1}+C_{2}} U_{o}\) (b) \(\frac{C_{1}}{C_{1}+C_{2}} U_{o}\) (c) \(\left(\frac{C_{1}-C_{2}}{C_{1}+C_{2}}\right)^{2} U_{o}\) (d) \(\frac{C_{1} C_{2}}{2\left(C_{1}+C_{2}\right)} U_{o}\)

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in (a) reduction of charge on the plates and increase of potential across the plates (b) increases in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates (c) decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates (d) none of the above

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