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Chapter 24: Problem 9

A parallel-plate air capacitor is to store charge of magnitude 240.0 \(\mathrm{pC}\) on each plate when the potential difference between the plates is 42.0 \(\mathrm{V}\) (a) If the area of each plate is \(6.80 \mathrm{cm}^{2},\) what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

Short Answer

Expert verified
(a) The separation is approximately 1.06 mm. (b) The new potential difference is about 84.2 V.

Step by step solution

01

Understand the Capacitance Formula

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

Calculate the Capacitance

First, we find the capacitance \( C \) using the relationship between charge \( Q \), capacitance \( C \), and voltage \( V \): \( Q = C \cdot V \). Solving for \( C \), we get \( C = \frac{Q}{V} = \frac{240.0 \times 10^{-12} \mathrm{C}}{42.0 \mathrm{V}} \approx 5.71 \times 10^{-12} \mathrm{F} \).
03

Calculate Separation Between the Plates

Rearrange the capacitance formula to solve for \( d \): \( d = \frac{\varepsilon_0 A}{C} \). Convert the area \( A = 6.80 \mathrm{cm}^2 \) to \( m^2 \) by dividing by 10000, so \( A = 6.80 \times 10^{-4} \mathrm{m}^2 \). Substitute the known values: \[ d = \frac{8.85 \times 10^{-12} \mathrm{F/m} \cdot 6.80 \times 10^{-4} \mathrm{m}^2}{5.71 \times 10^{-12} \mathrm{F}} \approx 1.06 \times 10^{-3} \mathrm{m} = 1.06 \mathrm{mm} \].
04

Calculate New Potential Difference for Double Separation

If the separation \( d \) is doubled, \( d' = 2 \times 1.06 \mathrm{mm} = 2.12 \mathrm{mm} = 2.12 \times 10^{-3} \mathrm{m} \). With the same charge, the new capacitance \( C' \) is \( \frac{\varepsilon_0 A}{d'} \). Substitute for \( d' \): \( C' = \frac{8.85 \times 10^{-12} \mathrm{F/m} \cdot 6.80 \times 10^{-4} \mathrm{m}^2}{2.12 \times 10^{-3} \mathrm{m}} \approx 2.85 \times 10^{-12} \mathrm{F} \). Use \( Q = C' \cdot V' \) to find the new voltage: \( V' = \frac{Q}{C'} = \frac{240.0 \times 10^{-12} \mathrm{C}}{2.85 \times 10^{-12} \mathrm{F}} \approx 84.2 \mathrm{V} \).

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Key Concepts

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

Capacitance Formula
Capacitance is a fundamental concept when studying capacitors, especially the parallel-plate type. To find the capacitance of a parallel-plate capacitor, we use the equation:
  • \( C = \frac{\varepsilon_0 A}{d} \)
Here, \( C \) is the capacitance, \( \varepsilon_0 \) (epsilon naught) is the permittivity of free space, \( A \) is the area of one of the plates, and \( d \) is the separation between the plates.
This formula shows that to increase capacitance you can either increase the plate area or reduce the separation between the plates. The permittivity of free space, a constant in this formula, represents how much electric field can permeate space, forming a crucial part of the calculation.
By rearranging this formula, if you already know the capacitance, area, and permittivity, you can solve for the separation \( d \) to understand more about the physical layout of your capacitor.
Potential Difference
Potential difference (voltage) in a capacitor is the work needed to move a charge between plates. It's closely tied with capacitance and charge storage via the formula:
  • \( Q = C \cdot V \)
This equation relates charge \( Q \) stored in the capacitor to its capacitance \( C \) and the potential difference \( V \) across its plates.
Simply put, the larger the potential difference, the more energy is stored per unit charge. It also means for a fixed charge, if you have a lower capacitance (such as from doubling the separation as in this exercise), the potential difference must be higher to keep the charge the same.
Understanding this interplay between these variables is crucial as it affects how capacitors are used in electrical circuits, deciding how much energy they can store or release under different conditions.
Permittivity of Free Space
One of the constants appearing in the formula for the capacitance calculation is the permittivity of free space, \( \varepsilon_0 \). It has a value of approximately \( 8.85 \times 10^{-12} \text{ F/m} \) (Farads per meter).
  • It's a measure of how easily an electric field can form in a vacuum.
The permittivity of free space sets the baseline for electrical permittivity, influencing how capacitors are made and how well they store charge.
It is essential in determining the capacitance of a parallel-plate capacitor because it acts as a proportionality factor linking the plate area and separation to the capacitance. In essence, when this value combines with the physical dimensions of the capacitor, it provides a peak into the potential effectiveness of the capacitor in storing electric energy.
Ensuring that you rightly understand and apply this constant is a critical step in the design and application of capacitors in electronic devices. It helps predict behavior under various electrical conditions.

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Most popular questions from this chapter

A \(5.80-\mu \mathrm{F}\), parallel-parallel-plate, air capacitor has a plate separation of 5.00 \(\mathrm{mm}\) and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3} .\)

An air capacitor is made by using two flat plates, each with area \(A,\) separated by a distance \(d\) . Then a a metal slab having thickness a (less than \(d\) ) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. P24.66). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance \(C_{0}\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a \rightarrow 0\) and \(a \rightarrow d\) .

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.0 \(\mathrm{pC}\) . The inner cylinder has radius \(0.50 \mathrm{mm},\) the outer one has radius \(5.00 \mathrm{mm},\) and the length of each cylinder is 18.0 \(\mathrm{cm} .\) (a) What is the capacitance? (b) What applied potential difference is necessary to produce these charges on the cylinders?

A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148\(\mu \mathrm{C}\) on each plate. The plates are 0.328 \(\mathrm{mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric- field magnitude between the plates? (d) What is the surface charge density on each plate?

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 sheets 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 \(12.0 \mathrm{mm},\) instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-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 Teflon than of poster board? Explain.
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