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24.20. Two parallel plate vacuum capacitors have plate spacings \(d_{1}\) and \(d_{2}\) and equal plate areas \(A\) . Show that when the capacitors are connected in series, the eqnivalent capacitance is the same as for a single capacitor with plate area \(A\) and spacing \(d_{1}+d_{2}\) .

Short Answer

Expert verified
The equivalent capacitance is \( \frac{\varepsilon_0 A}{d_1 + d_2} \), which matches a single capacitor with area \( A \) and spacing \( d_1 + d_2 \).

Step by step solution

01

Understanding Capacitance Formula

The capacitance for a parallel plate capacitor can be expressed as \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.
02

Express Individual Capacitances

For the first capacitor with plate separation \( d_1 \), the capacitance is \( C_1 = \frac{\varepsilon_0 A}{d_1} \). For the second capacitor with plate separation \( d_2 \), the capacitance is \( C_2 = \frac{\varepsilon_0 A}{d_2} \).
03

Capacitance in Series

For capacitors in series, the equivalent capacitance \( C_{eq} \) is given by \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \).
04

Calculate Equivalent Capacitance

Substituting the expressions for \( C_1 \) and \( C_2 \), we have:\[ \frac{1}{C_{eq}} = \frac{d_1}{\varepsilon_0 A} + \frac{d_2}{\varepsilon_0 A} \]Simplifying, \[ \frac{1}{C_{eq}} = \frac{d_1 + d_2}{\varepsilon_0 A} \]Thus, \[ C_{eq} = \frac{\varepsilon_0 A}{d_1 + d_2} \].
05

Comparison to Single Capacitor

This result shows that the equivalent capacitance \( C_{eq} \) is the same as for a single capacitor with the same plate area \( A \) and a total plate spacing of \( d_1 + d_2 \), as desired.

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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 consists of two flat, parallel conductive plates separated by a distance, with an insulating material, often a vacuum or dielectric, in between. It is one of the most straightforward setups for storing electrical energy.
  • The key components are the plate area, denoted by \( A \), and the distance between the plates, \( d \).
  • The ability of the capacitor to store charge is described by its capacitance \( C \), which for a parallel plate capacitor in a vacuum, is calculated using the formula \( C = \frac{\varepsilon_0 A}{d} \).
  • \( \varepsilon_0 \) represents the permittivity of free space, a constant value that determines how much electric field can be created per unit charge.
In our exercise, two capacitors with equal plate areas but different plate spacings are considered. When dealing with capacitors in series, understanding these basic concepts helps us analyze how their capacitances combine.
Equivalent Capacitance
When capacitors are connected in series, they function much like resistors, but the formula to compute their total or equivalent capacitance differs. For capacitors in series, the reciprocal of the equivalent capacitance \( C_{eq} \) is the sum of the reciprocals of the individual capacitances.
  • Expressed mathematically: \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \)
  • This means that the total equivalent capacitance will always be lower than the smallest individual capacitor in the series.
By substituting the values for the individual capacitance \( C_1 = \frac{\varepsilon_0 A}{d_1} \) and \( C_2 = \frac{\varepsilon_0 A}{d_2} \) in the given exercise, we arrive at the combined equivalent capacitance \( C_{eq} = \frac{\varepsilon_0 A}{d_1 + d_2} \). This demonstrates that the effect of capacitors in series involves an increase in effective plate spacing while maintaining the same plate area, thus reducing the overall capacitance.
Capacitance Formula
The capacitance formula for a parallel plate capacitor is foundational to understanding its behavior and function. It characterizes how effectively a capacitor can store charge based on physical properties of the capacitor itself.The formula is given by:\[ C = \frac{\varepsilon_0 A}{d} \]Where:- \( C \) stands for capacitance.- \( \varepsilon_0 \) is the permittivity constant indicating how material in between plates affects the capacitor's efficiency.- \( A \) is the area of the plates, larger areas can store more charge.- \( d \) is the distance between the plates, smaller distances increase capacitance since electrical field strength becomes larger.Understanding the capacitance formula in the context of our given problem, you can see that either increasing the area or decreasing the spacing increases capacitance. However, when capacitors are placed in series, as demonstrated, effective spacing increases to \( d_1 + d_2 \), which in turn lowers the equivalent capacitance, following the principles explained above.

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Most popular questions from this chapter

24.40. A budding electronics hobbyist wants to make a simple 1.0 -nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few shects of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 \(\mathrm{mm}\) . (a) If the sheets of paper measure \(22 \times 28 \mathrm{cm}\) and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of \(120 \mathrm{mm},\) instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 \(\mathrm{nF}\) of capacitance? (c) Suppose she goes hight-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Tellon than of posterboard? Explain.

24.5. A \(10.0-\mu \mathrm{F}\) paralle-plate capacitor with circular plates is connected to a 12.0 \(\mathrm{V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery?(c) How much charge would be on the plates if the capacitor were connected to the \(12.0-\mathrm{V}\) battery after the radius of each plate was doubled without changing their separation?

24.47. A \(12.5-\mu F\) capacitor is connected to a power supply that keeps a constant potential difference of 24.0 \(\mathrm{V}\) across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the diclectric is inserted? (b) By how much did the energy change dur- ing the insertion? Did it increase or decrease?

24.27. A \(450-\mu\) F capacitor is charged to 295 V. Then a wire is con- nected between the plates. How many joules of thermal energy are produced as the capacitor discharges if all of the energy that was stored goes into heating the wire?

24.58. Several \(0.25-\mu \mathrm{F}\) capacitors are available. The voltage across each is not to exceed 600 \(\mathrm{V}\) . You need to make a capacitor with capacitance 0.25\(\mu \mathrm{F}\) to be connected across a potential difference of 960 \(\mathrm{V}\) . (a) Show in a diagram how an cquivalent capacitor with the desired properties can be obtained. (b) No dielectric is a perfect insulator that would not permit the flow of any charge through its volume. Suppose that the dielectric in one of the capacitors in your diagram is a moderately good conductor. What will happen in this case when your combination of capacitors is connected across the \(960-\mathbf{V}\) potential difference?

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