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Chapter 10: Problem 5

Lightning discharges a cloud during an electrical storm. Suppose that the current in the lightning bolt varies with time as \(I=b t\), where \(b\) is a constant. Find the cloud's charge as a function of time.

Short Answer

Expert verified
The charge as a function of time is \(Q(t) = \frac{b}{2} t^2\).

Step by step solution

01

Understand the Relationship Between Current and Charge

The current, denoted as \(I(t)\), is the rate of change of charge \(Q(t)\) with respect to time. This relationship is expressed through the equation \(I(t) = \frac{dQ}{dt}\).
02

Substitute the Expression for Current

We are given that the current varies with time as \(I = bt\). So we substitute this into the rate of change of charge equation: \(bt = \frac{dQ}{dt}\). This means the rate at which charge accumulates is proportional to time.
03

Integrate to Find Charge as a Function of Time

To find the charge \(Q(t)\) as a function of time, integrate the equation \(bt = \frac{dQ}{dt}\) with respect to time. % \[ Q(t) = \int bt \ dt = \frac{b}{2} t^2 + C \]where \(C\) is the constant of integration, representing the initial charge at time zero.
04

Determine the Initial Condition (Optional)

If given an initial condition such as \(Q(0) = 0\), substitute \(t = 0\) into the expression to find \(C\). In this case, \(C = 0\), so the equation simplifies to \(Q(t) = \frac{b}{2} t^2\).

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Key Concepts

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

Understanding Current
Current is a fundamental concept in electricity and refers to the flow of electric charge. It is the rate at which charge flows through a conductor and is measured in amperes (A).
The formula to express current is simple:
  • Current (\( I \)) = Charge (\( Q \)) / Time (\( t \))
This shows that current (\( I \)) is the rate of flow of electric charge. If you think of electricity as a river, the current is the amount of water flowing past a point every second.
In our exercise, the current (\( I \)) is given as a function of time (\( t \)), specifically, \( I = bt \). Here, \( b \) is a constant that affects how rapidly the current increases over time.

By understanding how current varies, we can determine other aspects like how much charge flows over a specific time period.
Defining Electric Charge
Electric charge is a key property of particles that causes them to experience a force in an electromagnetic field. It's measured in coulombs (C) and can be positive or negative.
Charges are carried by electrons and protons:
  • Electrons carry a negative charge.
  • Protons carry a positive charge.
Charge is crucial as it forms the basis of electrical phenomena.
The relationship between charge (\( Q \)) and current (\( I \)) is given by the formula: \( I = \frac{dQ}{dt} \). This equation tells us the current is the rate at which charge changes over time.
In the exercise, to find the total charge \( Q(t) \) flowing through the cloud as a function of time, we used this relationship to integrate and determine the total charge based on the varying current.
Integration and Calculating Charge
Integration is a mathematical process used to find the total amount of something when given its rate of change. In the context of charge and current, integration allows us to determine the total charge accumulated over a period.
Given the equation for current \( I = bt \), the relationship with charge is described as \( \frac{dQ}{dt} = bt \).
To find the charge function \( Q(t) \), we integrate the current with respect to time:
  • \[ Q(t) = \int bt \ dt = \frac{b}{2} t^2 + C \]
Here, \( C \) represents the integration constant, which generally reflects the initial charge at the time measurement starts. If no initial conditions are provided and \( Q(0) = 0 \), then \( C \) is zero, simplifying our solution.
Integrating provides a powerful tool to transition from a rate of change (current) to accumulating the total quantity (charge) over time.

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Most popular questions from this chapter

Suppose six identical resistors, each with resistance \(R\), are connected so that they form the edges of a tetrahedron (a pyramid with three sides in addition to the base, i.e., one less side than an Egyptian pyramid). What resistance value or values can be obtained by making connections onto any two points on this arrangement? (solution in the pdf version of the book)

You have a circuit consisting of two unknown resistors in series, and a second circuit consisting of two unknown resistors in parallel. (a) What, if anything, would you learn about the resistors in the series circuit by finding that the currents through them were equal? (b) What if you found out the voltage differences across the resistors in the series circuit were equal? (c) What would you learn about the resistors in the parallel circuit from knowing that the currents were equal? (d) What if the voltages in the parallel circuit were equal?

(a) Recall that the gravitational energy of two gravitationally interacting spheres is given by \(P E=-G m_{1} m_{2} / r\), where \(r\) is the center-to- center distance. Sketch a graph of \(P E\) as a function of \(r\), making sure that your graph behaves properly at small values of \(r\), where you're dividing by a small number, and at large ones, where you're dividing by a large one. Check that your graph behaves properly when a rock is dropped from a larger \(r\) to a smaller one; the rock should lose potential energy as it gains kinetic energy. (b) Electrical forces are closely analogous to gravitational ones, since both depend on \(1 / r^{2}\). Since the forces are analogous, the potential energies should also behave analogously. Using this analogy, write down the expression for the electrical potential energy of two interacting charged particles. The main uncertainty here is the sign out in front. Like masses attract, but like charges repel. To figure out whether you have the right sign in your equation, sketch graphs in the case where both charges are positive, and also in the case where one is positive and one negative; make sure that in both cases, when the charges are released near one another, their motion causes them to lose PE while gaining KE.(answer check available at lightandmatter.com)

You are given a battery, a flashlight bulb, and a single piece of wire. Draw at least two configurations of these items that would result in lighting up the bulb, and at least two that would not light it. (Don't draw schematics.) If you're not sure what's going on, borrow the materials from your instructor and try it. Note that the bulb has two electrical contacts: one is the threaded metal jacket, and the other is the tip (at the bottom in the figure). [Problem by Arnold Arons.]

If a typical light bulb draws about \(900 \mathrm{~mA}\) from a 110-V household circuit, what is its resistance? (Don't worry about the fact that it's alternating current.) (answer check available at lightandmatter.com)
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