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To make a \(0.40-\mu F\) capacitor, what area must the plates have if they are to be separated by a 28 -mm air gap?

Short Answer

Expert verified
The plates need an area of approximately \(1.27 \, \text{m}^2\).

Step by step solution

01

Understand the formula for capacitance

The capacitance for a parallel plate capacitor is given by the formula:\[ C = \frac{\varepsilon_0 \cdot A}{d} \]where \(C\) is the capacitance, \(\varepsilon_0\) is the permittivity of free space \( (8.85 \times 10^{-12} \, \text{F/m}) \), \(A\) is the area of one of the plates in square meters, and \(d\) is the separation between the plates in meters.
02

Plug in the given values

We know the capacitance \(C = 0.40 \, \mu \text{F} = 0.40 \times 10^{-6} \, \text{F}\) and the separation \(d = 28\, \text{mm} = 28 \times 10^{-3} \, \text{m} \). We need to solve for \(A\).
03

Rearrange the formula to solve for area

Rearrange the formula to solve for \(A\):\[ A = \frac{C \cdot d}{\varepsilon_0} \]
04

Calculate the area of the plates

Plug in the known values into the rearranged equation:\[ A = \frac{0.40 \times 10^{-6} \, \text{F} \times 28 \times 10^{-3} \, \text{m}}{8.85 \times 10^{-12} \, \text{F/m}} \]Calculate the area to find that:\[ A \approx 1.27 \, \text{m}^2 \]
05

Provide the result

The area of the plates must be approximately \(1.27 \, \text{m}^2\).

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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 one of the simplest types of capacitors, consisting of two conductive plates separated by an insulating material, known as a dielectric. It's instrumental in storing electrical energy by maintaining a charge. The effectiveness of a capacitor largely depends on its construction. When a voltage is applied across the plates, an electric field is created, allowing for the accumulation of charge: one plate holds positive charge, while the other holds an equal amount of negative charge. Unlike other capacitors, the defining feature of parallel plate capacitors is their structural simplicity and predictability based on physical dimensions. The capacitance, or the ability to store charge, hinges on two primary factors:
  • The surface area of the plates: Larger plates can store more charge simply because they provide more space for charges to reside.
  • The distance between the plates, known as the separation: Shorter distances generally yield a higher capacitance because the electric field strength is more concentrated.
Parallel plate capacitors are a popular choice in many electronic devices due to their straightforward design and reliable performance.
Permittivity of Free Space
The permittivity of free space, denoted as \(\varepsilon_0\), plays a crucial role in the world of electromagnetism, especially concerning capacitors. It represents the ability of the vacuum to permit electric field lines, acting essentially like a "permittivity constant" for free space. This constant reflects how electric fields interact with the space around them.Its value is \(8.85 \times 10^{-12} \, \text{F/m}\), and it's a foundational constant used in the formula for capacitance of capacitors, particularly those involving parallel plates. In the formula, \(C = \frac{\varepsilon_0 \cdot A}{d}\), \(\varepsilon_0\) quantifies how much electric flux can "flow" through a vacuum. In practical terms, when calculating capacitance, \(\varepsilon_0\) serves as a scaling factor, ensuring that the relationship between plate area, separation distance, and stored charge is accurately represented.It's crucial to understand that while materials between capacitor plates can influence capacitance significantly, when calculating in a vacuum or air, \(\varepsilon_0\) is your go-to constant. This universality makes it an essential component in physics and engineering studies.
Capacitor Area Calculation
Calculating the area of the plates in a capacitor is fundamentally tied to the formula for capacitance: \(C = \frac{\varepsilon_0 \cdot A}{d} \,\). In this formula, \(A\) represents the area of one of the plates in square meters. Given the capacitance \(C\), plate separation \(d\), and the permittivity of free space \(\varepsilon_0\), one can rearrange this equation to find the area necessary.To solve for \(A\), the operation becomes:\[A = \frac{C \cdot d}{\varepsilon_0}\]This rearrangement is essential because it's straightforward to measure or specify capacitance and distance, but the area might be more challenging to determine without this relationship.**Example Calculation:** Suppose you have a capacitance of \(0.40 \, \mu F\) and the plates are separated by \(28 \, \text{mm} = 28 \times 10^{-3} \, \text{m}\). Plug the known values into the equation for \(A\):\[A = \frac{0.40 \times 10^{-6} \, F \times 28 \times 10^{-3} \, m}{8.85 \times 10^{-12} \, F/m}\]This calculation gives an area \(\approx 1.27 \, \text{m}^2\).Understanding this calculation allows engineers and designers to optimize capacitor designs efficiently, balancing size and performance to best meet their needs. It also emphasizes the importance of both correctly measured and theoretically calculated values in practical applications.

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Most popular questions from this chapter

A \(0.50-\mu \mathrm{F}\) and a \(0.80-\mu \mathrm{F}\) capacitor are connected in series to a 9.0-V battery. Calculate (a) the potential difference across each capacitor and \((b)\) the charge on each. (c) Repeat parts \((a)\) and \((b)\) assuming the two capacitors are in parallel.

A parallel-plate capacitor with plate area \(2.0 \mathrm{~cm}^{2}\) and airgap separation \(0.50 \mathrm{~mm}\) is connected to a \(12-\mathrm{V}\) battery, and fully charged. The battery is then disconnected. ( \(a\) ) What is the charge on the capacitor? (b) The plates are now pulled to a separation of \(0.75 \mathrm{~mm}\). What is the charge on the capacitor now? (c) What is the potential difference across the plates now? ( \(d\) ) How much work was required to pull the plates to their new separation?

In an electrostatic air cleaner ( "precipitator"), the strong nonuniform electric field in the central region of a cylindrical capacitor (with outer and inner cylindrical radii \(R_{\mathrm{a}}\) and \(R_{\mathrm{b}}\) ) is used to create ionized air molecules for use in charging dust and soot particles (Fig. \(24-21\) ). Under standard atmospheric conditions, if air is subjected to an electric field magnitude that exceeds its dielectric strength \(E_{\mathrm{S}}=2.7 \times 10^{6} \mathrm{~N} / \mathrm{C},\) air molecules will dissociate into positively charged ions and free electrons. In a precipitator, the region within which air is ionized (the corona discharge region) occupies a cylindrical volume of radius \(R\) that is typically five times that of the inner cylinder. Assume a particular precipitator is constructed with \(R_{\mathrm{b}}=0.10 \mathrm{~mm}\) and \(R_{\mathrm{a}}=10.0 \mathrm{~cm} .\) In order to create a corona discharge region with radius \(R=5.0 \mathrm{R}_{\mathrm{b}},\) what potential difference \(V\) should be applied between the precipitator's inner and outer conducting cylinders? [Besides dissociating air, the charged inner cylinder repels the resulting positive ions from the corona discharge region, where they are put to use in charging dust particles, which are then "collected" on the negatively charged outer cylinder.]

A parallel-plate capacitor is isolated with a charge \(\pm Q\) on each plate. If the separation of the plates is halved and a dielectric (constant \(K\) ) is inserted in place of air, by what factor does the energy storage change? To what do you attribute the change in stored potential energy? How does the new value of the electric field between the plates compare with the original value?

A parallel-plate capacitor has square plates 12 \(\mathrm{cm}\) on a side separated by 0.10 \(\mathrm{mm}\) of plastic with a dielectric constant of \(K=3.1 .\) The plates are connected to a battery, causing them to become oppositely charged. Since the oppositely charged plates attract each other, they exert a pressure on the dielectric. If this pressure is 40.0 \(\mathrm{Pa}\) , what is the battery voltage?

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