/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Calculate the work done by a \(3... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The work done is 35.1 µJ.

Step by step solution

01

Understand the relationship between charge, capacitance, and voltage

The work done by the battery is equivalent to the energy stored in the capacitor as it charges. The energy (or work done) stored in a capacitor can be calculated using the formula \( W = \frac{1}{2} CV^2 \), where \( W \) is the work done in joules, \( C \) is the capacitance in farads, and \( V \) is the voltage in volts. In this problem, the capacitance \( C \) is given as \(7.8\,\mu\mathrm{F} = 7.8 \times 10^{-6}\,\mathrm{F}\) and the voltage \( V \) is \(3.0\,\mathrm{V}\).
02

Substitute the given values into the formula

Substitute the values for capacitance and voltage into the formula: \[ W = \frac{1}{2} \times 7.8 \times 10^{-6} \times (3.0)^2 \]
03

Calculate the work done

First, calculate the square of the voltage: \[ (3.0)^2 = 9.0 \]Then, substitute this value back into the equation:\[ W = \frac{1}{2} \times 7.8 \times 10^{-6} \times 9.0 \]Calculate the product:\[ W = \frac{1}{2} \times 70.2 \times 10^{-6} \]\[ W = 35.1 \times 10^{-6} \]The value of \( W \) is \(35.1 \times 10^{-6}\,\mathrm{J} \) or \(35.1\,\mu\mathrm{J} \).

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Key Concepts

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

energy stored in capacitors
When we talk about capacitors, one of the key concepts is the energy they store. Capacitors are electrical components capable of holding a charge temporarily, similar to how a battery stores energy. As a capacitor charges, it stores energy in its electric field. This energy can later be released to do work. The energy stored in a capacitor can be calculated using the formula:\[ W = \frac{1}{2} CV^2 \]Where:
  • \( W \) is the work or energy stored in joules (J)
  • \( C \) is the capacitance, measured in farads (F)
  • \( V \) is the voltage across the capacitor, measured in volts (V)
When a capacitor charges, the energy increase follows an exponential curve, meaning it charges rapidly at first and then gradually slows down until reaching its maximum stored energy. This stored energy is crucial for applications like camera flashes, where a sudden, large burst of energy is needed.
capacitance and voltage relationship
Understanding the relationship between capacitance and voltage is key to effectively using capacitors. Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It depends on the physical characteristics of the capacitor, such as the surface area of the plates and the distance between them, along with the dielectric material in between. This relationship is expressed in the equation:\[ Q = CV \]Where:
  • \( Q \) is the charge stored in coulombs (C)
  • \( C \) is the capacitance in farads (F)
  • \( V \) is the voltage in volts (V)
This equation highlights how an increase in voltage across a capacitor causes a linear increase in charge stored, assuming the capacitance remains constant. When applying a constant voltage, a larger capacitance means more charge storage. This relationship is fundamental when designing circuits to ensure they function as intended, especially in timing and pulse applications.
charging a capacitor
Charging a capacitor involves the transfer of energy from a power source, such as a battery, to the capacitor. This process occurs until the voltage across the capacitor equals the voltage of the battery. Initially, the current flow is maximum when the charging process starts. Over time, as the capacitor accumulates charge, the current decreases.During the charging, the voltage across the capacitor increases and follows an exponential curve. Eventually, it approaches but never quite reaches the full battery voltage, practically speaking. This exponential curve emphasizes that most of the charging happens fairly quickly, but reaching complete charge takes time.A key formula related to charging is derived from Ohm's law in a simple RC (resistor-capacitor) circuit:\[ V(t) = V_0 (1 - e^{-t/RC}) \]Where:
  • \( V(t) \) is the voltage at time \( t \)
  • \( V_0 \) is the initial voltage
  • \( R \) is the resistance in ohms (Ω)
  • \( C \) is the capacitance in farads (F)
  • \( t \) is the time in seconds
  • \( e \) is the base of the natural logarithm
Understanding how a capacitor charges is vital in electronics, where precise timing is necessary, such as in oscillators or filter circuits.

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Most popular questions from this chapter

A parallel-plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the spacing between the plates is doubled, how is the magnitude of charge on the plates affected?

Find the potential difference required to accelerate protons from rest to \(10 \%\) of the speed of light. (At this point, relativistic effects start to become significant)

Rutherford's Planetary Model of the Atom In 1911 , Ernest Rutherford developed a planetary model of the atom, in which a small positively charged nucleus is orbited by electrons. The model was motivated by an experiment carried out by Rutherford and his graduate students, Geiger and Marsden. In this experiment, they fired alpha particles with an initial speed of \(1.75 \times 10^{7} \mathrm{m} / \mathrm{s}\) at a thin sheet of gold. (Alpha particles are obtained from certain radioactive decays. They have a charge of \(+2 e\) and a mass of \(6.64 \times 10^{-27} \mathrm{kg}\).) How close can the alpha particles get to a gold nucleus (charge \(=+79 e\) ), assuming the nucleus remains stationary? (This calculation sets an upper limit on the size of the gold nucleus. See Chapter 31 for further details.)

Cell Membranes and Dielectrics Many cells in the body have a cell membrane whose inner and outer surfaces carry opposite charges, just like the plates of a parallel-plate capacitor. Suppose a typical cell membrane has a thickness of \(8.1 \times 10^{-9} \mathrm{m},\) and its inner and outer surfaces carry charge densities of \(-0.58 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2}\) and \(+0.58 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) respectively. In addition, assume that the material in the cell membrane has a dielectric constant of \(5.5 .\) (a) Find the direction and magnitude of the electric field within the cell membrane. (b) Calculate the potential difference between the inner and outer walls of the membrane, and indicate which wall of the membrane has the higher potential.

In the Bohr model of the hydrogen atom, a proton and an electron are separated by a constant distance \(r .\) (a) Would the electric potential energy of the system increase, decrease, or stay the same if the electron is replaced with a proton? Explain. (b) Suppose, instead, that the proton is replaced with an electron. Would the electric potential energy of the system increase, decrease, or stay the same? Explain.

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