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Chapter 24: Problem 31

24.31. (a) How much charge does a battery have to supply to a 5.0 - \(\mu \mathrm{F}\) capacitor to create a potential difference of 1.5 \(\mathrm{V}\) across its plates? How much energy is stored in the capacitor in this case? (b) How much charge would the battery have to supply to store 1.0 \(\mathrm{J}\) of energy in the capacitor? What would be the potential across the capacitor in that case?

Short Answer

Expert verified
(a) The charge is \( 7.5 \times 10^{-6} \mathrm{C} \) and the energy is \( 5.625 \times 10^{-6} \mathrm{J} \). (b) The charge is \( 3.162 \times 10^{-3} \mathrm{C} \) and the potential is \( 632.4 \mathrm{V} \).

Step by step solution

01

Understand the Problem

The problem requires calculating charge and energy for a capacitor based on given parameters: capacitance and potential difference (voltage). There are two parts: (a) uses the given voltage to find charge and energy, while (b) requires finding charge and voltage to achieve a specific energy storage in the capacitor.
02

Find Charge with Given Voltage (Part a)

Use the formula for charge stored in a capacitor, \[ Q = C \cdot V \]where \( C = 5.0 \cdot 10^{-6} \mathrm{F} \) (capacitor value) and \( V = 1.5 \mathrm{V} \). Calculate: \[ Q = 5.0 \cdot 10^{-6} \cdot 1.5 = 7.5 \cdot 10^{-6} \mathrm{C} \] So, the battery must supply \( 7.5 \cdot 10^{-6} \mathrm{C} \) of charge.
03

Calculate Energy Stored (Part a)

Use the energy formula for a capacitor,\[ U = \frac{1}{2} C V^2 \]Substitute values \( C = 5.0 \cdot 10^{-6} \mathrm{F} \) and \( V = 1.5 \mathrm{V} \):\[ U = \frac{1}{2} \times 5.0 \cdot 10^{-6} \times (1.5)^2 = 5.625 \times 10^{-6} \mathrm{J} \]Thus, the energy stored is \( 5.625 \times 10^{-6} \mathrm{J} \).
04

Find Charge with Given Energy (Part b)

Use the energy formula again, but rearrange it to find charge \[ U = \frac{1}{2} \cdot \frac{Q^2}{C} \] Thus, \( Q^2 = 2U \cdot C \). Substitute \( U = 1.0 \mathrm{J} \) and \( C = 5.0 \cdot 10^{-6} \mathrm{F} \): \[ Q^2 = 2 \times 1.0 \times 5.0 \cdot 10^{-6} \rightarrow Q = \sqrt{1.0 \times 10^{-5}} = 0.003162 \mathrm{C} \] So, the charge supplied is approximately \( 3.162 \times 10^{-3} \mathrm{C} \).
05

Calculate Potential for Given Energy (Part b)

Substitute the values into the charge formula \( Q = C \cdot V \), solving for voltage:\[ V = \frac{Q}{C} = \frac{0.003162}{5.0 \cdot 10^{-6}} = 632.4 \mathrm{V} \]Therefore, the potential difference needed is approximately \( 632.4 \mathrm{V} \).

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Key Concepts

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

Electrostatics
Electrostatics is the branch of physics that studies stationary or slow-moving electric charges. It's a fundamental component of many electrical and electronic devices, such as capacitors. A capacitor stores electric energy by maintaining a static charge on its conductive plates. The potential difference, often referred to as voltage, between these plates is what attracts the charges. A significant aspect of electrostatics is the concept of electric fields. These are fields around a charged object, influencing other charges in the vicinity.

Key points to understand about electrostatics in capacitors include:
  • Capacitors consist of two conductive plates separated by an insulating material.
  • The electric field in a capacitor is typically uniform between the plates.
  • Electrostatic forces are responsible for moving charges from one plate to the other under the influence of voltage.
Such concepts are crucial in understanding not only capacitors but also the broader implications of electrostatic principles in physics.
Capacitance
Capacitance is a measure of a capacitor's ability to store electric charge. It is denoted by the letter "C" and is typically measured in farads (F). The basic formula that relates the charge (Q), capacitance (C), and potential difference (V) is given by:\[ Q = C \cdot V \]This equation shows how capacitance influences the amount of charge a capacitor can hold at a certain voltage.

For practical use:
  • A higher capacitance means the capacitor can store more charge at a given voltage.
  • Capacitance depends on physical factors such as the area of the plates and the distance between them.
  • Materials used as dielectrics (the insulating part between plates) significantly affect the capacitance.
Understanding capacitance is essential for designing circuits involving energy storage, filtering, timing elements, and much more.
Energy Storage in Capacitors
Storing energy in capacitors is one of their primary functions. The formula used to calculate the energy (U) stored in a capacitor is:\[ U = \frac{1}{2} C V^2 \]This formula highlights the dependence of stored energy on both the capacitance of the capacitor and the square of the voltage applied across it.

Considerations about energy storage:
  • The energy stored is doubled if the voltage is doubled, emphasizing the relation of energy with the square of voltage.
  • For applications requiring immediate energy bursts, capacitors are advantageous thanks to their rapid charge and discharge cycles.
  • Capacitors can only store a fraction of the energy that a battery can, but they release energy much faster.
Understanding how capacitors store and release energy is fundamental in power supply applications, transient voltage stabilization, and signal processing devices.

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Most popular questions from this chapter

24.65. A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 \(\mathrm{V}\) when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 \(\mathrm{V}\) . What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

24.48 . A parallel-plate capacitor has plates with area 0.0225 \(\mathrm{m}^{2}\) separated by 1.00 \(\mathrm{mm}\) of Tefion. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 \(\mathrm{V}\) . (b) Use Gauss's law (Eq. 24.23 ) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

24.50. A parallel-plate air capacitor is made by using two plates 16 \(\mathrm{cm}\) square, spaced 4.7 \(\mathrm{mm}\) apart. It is connected to a \(12-\mathrm{V}\) bat- tery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of \(9.4 \mathrm{mm},\) what are the answers to parts \((\mathrm{a})-(\mathrm{d}) ?\)

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?

24.9. 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?
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