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Chapter 20: Problem 52

To operate a given flash lamp requires a charge of \(32 \mu C .\) What capacitance is needed to store this much charge in a capacitor with a potential difference between its plates of \(9.0 \mathrm{V} ?\)

Short Answer

Expert verified
The required capacitance is 3.56 渭F.

Step by step solution

01

Identify the Formula

To find capacitance, use the formula for capacitance: \[ C = \frac{Q}{V} \]where \( C \) is capacitance, \( Q \) is charge, and \( V \) is the potential difference.
02

Substitute Given Values

Substitute the given values into the formula. We have \( Q = 32 \mu C = 32 \times 10^{-6} C \) and \( V = 9.0 \, V \). So the formula becomes:\[ C = \frac{32 \times 10^{-6}}{9.0} \]
03

Calculate Capacitance

Perform the calculation:\[ C = \frac{32 \times 10^{-6}}{9.0} = 3.56 \times 10^{-6} F \]So, the required capacitance is \( 3.56 \mu F \).

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Key Concepts

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

Charge
Charge is the foundational concept of electric fields and circuits. It refers to the amount of electric energy stored in an object like a capacitor.
It is measured in Coulombs, represented by the symbol \(Q\). Here, the charge value provided is \(32 \, \mu C\), where \(\mu C\) stands for microcoulombs, equivalent to \(32 \times 10^{-6} \, C\). This small unit is commonly used in electronics due to the typically tiny charges involved in circuits.
Understanding charge is crucial:
  • Charge is either positive or negative鈥攍ike forces repel, opposites attract.
  • It is conserved within a closed system, so the total remains constant.
  • Charge can be transferred from one object to another, creating electrical potential.
In the context of a flash lamp, the charge determines how much energy can be stored and released to produce light.
Potential Difference
Potential difference, often referred to as voltage, is the driving force that pushes electric charge through a circuit. It is measured in volts (V) and is symbolized by \(V\). For the given example, the potential difference between the capacitor plates is \(9.0 \, V\).
Key aspects of potential difference:
  • It defines the energy per unit charge, measured as \(1 \, Joule/Coulomb\).
  • A higher potential difference indicates a stronger "push" on the charge, facilitating current flow.
  • It is the reason electric circuits can perform work, such as lighting a flash lamp.
In capacitors, the potential difference is harnessed to store and release energy efficiently. Understanding how this relates to charge helps explain how devices like cameras or electronic flashes store energy for use.
Capacitors
Capacitors are devices that store electrical energy in an electric field, consisting of two conductive plates separated by an insulator. They play a key role in electronic circuits, powering devices when needed.
Understanding capacitors:
  • They temporarily hold electric charge, releasing it when the circuit needs power.
  • Their ability to store charge is called "capacitance," denoted by \(C\), measured in farads (F).
  • The formula \[ C = \frac{Q}{V} \] expresses capacitance, showing the relationship between charge (Q), potential difference (V), and capacitance (C).
In the exercise, given a charge of \(32 \, \mu C\) and a potential difference of \(9.0 \, V\), the capacitance calculated is \(3.56 \, \mu F\). This shows the capacitor鈥檚 capacity to store and manage electrical energy efficiently. Capacitors facilitate various functions, from smoothing voltage in circuits to triggering flashes in cameras.

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Most popular questions from this chapter

Computer Keyboards Many computer keyboards operate on the principle of capacitance. As shown in Figure \(20-16\) each key forms a small parallel-plate capacitor whose separation is reduced when the key is depressed. (a) Does depressing a key increase or decrease its capacitance? Explain. (b) Suppose the plates for each key have an area of \(47.5 \mathrm{mm}^{2}\) and an initial separation of \(0.550 \mathrm{mm}\). In addition, let the dielectric have a dielectric constant of \(3.75 .\) If the circuitry of the computer can detect a change in capacitance of \(0.425 \mathrm{pF}\), what is the minimum distance a key must be depressed to be detected?

A point charge of \(-7.2 \mu C\) is at the origin. What is the electric potential at (a) \((3.0 \mathrm{m}, 0)\) (b) \((-3.0 \mathrm{m}, 0) ;\) and (c) \((3.0 \mathrm{m},-3.0 \mathrm{m}) ?\)

Defibrillator An automatic external defibrillator (AED) delivers 125 J of energy at a voltage of 1050 V. What is the capacitance of this device?

A parallel-plate capacitor has plates with an area of \(0.012 \mathrm{m}^{2}\) and a separation of \(0.88 \mathrm{mm} .\) The space between the plates is filled with a dielectric whose dielectric constant is \(2.0 .\) (a) What is the potential difference between the plates when the charge on the capacitor plates is \(4.7 \mu \mathrm{C}\) ? (b) Will your answer to part (a) increase, decrease, or stay the same if the dielectric constant is increased? Explain. (c) Calculate the potential difference for the case where the dielectric constant is \(4.0 .\)

Calculate the work done by a \(3.0-\mathrm{V}\) battery as it charges a 7.8- \(\mu \mathrm{F}\) capacitor in the flash unit of a camera.
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