/*! 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 26 A 60.0-pF vacuum capacitor has a... [FREE SOLUTION] | 91Ó°ÊÓ

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A 60.0-pF vacuum capacitor has a plate area of \(0.010 \mathrm{m}^{2} .\) What is the separation between its plates?

Short Answer

Expert verified
The separation between the plates in the vacuum capacitor is approximately \(1.48\,\mathrm{mm}\).

Step by step solution

01

Identify the given values

: We are given: - Capacitance (C) = 60.0 pF (Remember! 1 pF = \(10^{-12}\) F) - Plate area (A) = 0.010 m² The vacuum permittivity constant, \(\epsilon_0\), is a known value, which is approximately equal to \(8.854 × 10^{-12} \mathrm{F/m}\).
02

Convert Capacitance to SI units

: We'll first convert the given capacitance from pF to F: \[C = 60.0 \mathrm{\,pF} = 60.0 × 10^{-12} \mathrm{\,F}\]
03

Rearrange the formula for d

: Now, let's rearrange the capacitance formula to get an expression for the plate separation (d): \[d = \dfrac{\epsilon_0 × A}{C}\]
04

Substitute the given values and find d

: Now, let's substitute the given values and the vacuum permittivity constant into the formula derived in the previous step: \[d = \dfrac{8.854 × 10^{-12} \mathrm{F/m} × 0.010 \mathrm{\,m}^2}{60.0 × 10^{-12} \mathrm{\,F}}\] After calculating the numerator and denominator, we'll get: \[d = \dfrac{8.854 × 10^{-14} \mathrm{\,F\,m}}{60.0 × 10^{-12}\mathrm{\,F}}\] Finally, dividing the numerator by the denominator, we obtain the plate separation: \[d = 1.476 × 10^{-3} \mathrm{\,m}\] or \[d \approx 1.48\,\mathrm{mm}\] So, the separation between the plates in the vacuum capacitor is approximately 1.48 mm.

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Key Concepts

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

Vacuum Capacitor
A vacuum capacitor is a type of capacitor where the space between its plates is filled with a perfect vacuum. This is an important concept because vacuum capacitors have no dielectric material other than vacuum, which means their behavior is determined solely by their physical structure and the vacuum permittivity. These capacitors can handle high voltages, making them useful in applications such as radio transmitters and television transmitters.

The main advantage of using a vacuum capacitor is that it eliminates losses introduced by dielectric materials. In a vacuum capacitor, there are two plates separated by a distance with no material in between, providing high efficiency. Understanding the basics of vacuum capacitors is essential for those studying electronics and electromagnetic fields.
Capacitance Calculation
Capacitance is the ability of a system to store charge per unit voltage applied. The formula for calculating the capacitance of a parallel plate capacitor is given by: \[ C = \frac{\epsilon_0 \cdot A}{d} \] where:
  • C is the capacitance,
  • \(\epsilon_0\) is the vacuum permittivity constant, approximately \(8.854 \times 10^{-12} \mathrm{F/m}\),
  • A is the area of one of the plates,
  • d is the separation between the plates.
The capacitance increases proportionally with the plate area and decreases with the increase in plate separation. Therefore, larger plate areas or smaller separations create capacitors with higher capacitance.

In practice, to solve problems relating to capacitance, you identify given values, convert them to the correct units if necessary, and plug them into the formula. Understanding this process is crucial for designing and analyzing circuits in electronics.
Dielectric Constant
The dielectric constant, also called the relative permittivity, is a measure of a material's ability to increase the capacitance compared to a vacuum. However, in a vacuum capacitor, there is no dielectric other than the vacuum itself, which has a dielectric constant of 1.

The presence of a dielectric material would normally enhance the capacitor's ability to store charge due to its property of polarizing in an electric field, thereby reducing the electric field strength within the material itself. Even though vacuum itself does not provide this benefit, understanding the concept of dielectric constant is significant because it plays a crucial role when dealing with capacitors filled with other materials besides vacuum.

In exercises focusing on vacuum capacitors, the lack of dielectric material simplifies the situation, making it easier to grasp the essential relationship between capacitance, plate area, and plate separation.

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Most popular questions from this chapter

Suppose you need a capacitor bank with a total capacitance of 0.750 F but you have only 1.50 -mF capacitors at your disposal. What is the smallest number of capacitors you could connect together to achieve your goal, and how would you connect them?

For a Teflon \(^{\mathrm{TM}}\)-filled, parallel-plate capacitor, the area of the plate is \(50.0 \mathrm{cm}^{2}\) and the spacing between the plates is \(0.50 \mathrm{mm}\). If the capacitor is connected to a 200 -V battery, find (a) the free charge on the capacitor plates, (b) the electrical field in the dielectric, and (c) the induced charge on the dielectric surfaces.

An air-filled (empty) parallel-plate capacitor is made from two square plates that are \(25 \mathrm{cm}\) on each side and 1.0 mm apart. The capacitor is connected to a 50 -V battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of \(2.00 \mathrm{mm}\). (a) What is the capacitance of this new capacitor? (b) What is the charge on each plate? (c) What is the electrical field between the plates?

A spherical capacitor is formed from two concentric spherical conducting shells separated by a vacuum. The inner sphere has radius \(12.5 \mathrm{cm}\) and the outer sphere has radius \(14.8 \mathrm{cm} .\) A potential difference of \(120 \mathrm{V}\) is applied to the capacitor. (a) What is the energy density at \(r=12.6 \mathrm{cm},\) just outside the inner sphere? (b) What is the energy density at \(r=14.7 \mathrm{cm},\) just inside the outer sphere? (c) For the parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for the spherical capacitor?

Suppose that the capacitance of a variable capacitor can be manually changed from 100 to 800 pF by turning a dial connected to one set of plates by a shaft, from \(0^{\circ}\) to \(180^{\circ}\). With the dial set at \(180^{\circ}\) (corresponding to \(C=800 \mathrm{pF}),\) the capacitor is connected to a \(500-\mathrm{V}\) source. After charging, the capacitor is disconnected from the source, and the dial is turned to \(0^{\circ} .\) (a) What is the charge on the capacitor? (b) What is the voltage across the capacitor when the dial is set to \(0^{\circ} ?\)

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