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Chapter 1: Problem 5

The current flowing into a conductor varics in time according to the equation \(l=I_{0} e^{-\alpha !}\), where \(I_{0}\) and \(\alpha\) are constants. Find the charge \(Q\) which has accumulated on the conductor after a time \(t_{0}\). Sketch the variation of \(I\) and \(Q\) with time.

Short Answer

Expert verified
Answer: The accumulated charge \(Q(t_{0})\) at time \(t_{0}\) is given by the expression \(Q(t_{0}) = (I_{0}/\alpha) (1 - e^{-\alpha t_{0}})\).

Step by step solution

01

Rewrite the given equation for current

The current flowing into the conductor is given by the equation \(l=I_{0} e^{-\alpha t}\). Let's denote the current with \(I\), so we have: \(I = I_{0} e^{-\alpha t}\) We will work with this equation in the next steps to find the accumulated charge.
02

Integrate the current equation with respect to time

To find the charge (\(Q\)) after a time \(t_{0}\), we need to integrate the current over time. Recall that the charge is the integral of current. Therefore, we have: \(Q(t) = \int_{0}^{t} I dt = \int_{0}^{t} I_{0} e^{-\alpha t} dt\)
03

Evaluate the integral

Now, we will evaluate the integral: \(Q(t) = I_{0} \int_{0}^{t} e^{-\alpha t} dt\) Applying integration by substitution, let -\(u = \alpha t\) and du/dt = -\(\alpha\), so dt = du / (-\(\alpha\)). The new integral becomes: \(Q(t) = I_{0} \int_{0}^{-\alpha t} e^{u} (-\frac{1}{\alpha}) du\) Integrate and simplify: \(Q(t) = -(I_{0}/\alpha) \int_{0}^{-\alpha t} e^{u} du\) \(Q(t) = (I_{0}/\alpha) \left[e^{u} \right]_{0}^{-\alpha t}\) \(Q(t) = (I_{0}/\alpha) (e^{0} - e^{-\alpha t})\)
04

Find the charge at time \(t_{0}\)

To find the accumulated charge at time \(t_{0}\), we can simply substitute \(t = t_{0}\) into the equation for \(Q(t)\): \(Q(t_{0}) = (I_{0}/\alpha) (1 - e^{-\alpha t_{0}})\)
05

Sketch the variation of I and Q with time

To sketch the variation of I and Q with time, we should first note the initial conditions: - At \(t = 0\), \(I = I_{0}\) and \(Q = 0\) - As \(t\) increases, the exponential term in both \(I\) and \(Q\) will decay. Now, sketch both equations (\(I = I_{0} e^{-\alpha t}\) and \(Q = (I_{0}/\alpha) (1 - e^{-\alpha t})\)) separately on y-axis (with \(I\) and \(Q\)) and x-axis (with time \(t\)). The current \(I\) will start at \(I_{0}\) and decay exponentially with time towards 0. The charge \(Q\) will start at 0 and increase asymptotically towards \(\frac{I_{0}}{\alpha}\) as time approaches infinity.

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Key Concepts

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

Exponential Decay in Electrical Circuits
In electrical contexts, exponential decay describes how a current or voltage decreases over time at a rate proportional to its current value. For instance, in a circuit with a capacitor discharging through a resistor, the voltage across the capacitor or the current in the circuit exhibits exponential decay.

The equation for a time-varying current that decreases exponentially is often written as \( I = I_{0} e^{-\beta t} \), where \(I_{0}\) represents the initial current at time \(t = 0\), \(e\) is the base of the natural logarithm, \(\beta\) is the decay constant that determines how fast the current decays, and \(t\) is the time. The negative sign in the exponent indicates the decrease over time.

As time goes on, the current's value approaches zero, which is graphically represented as a curve that swiftly drops from the initial current level and gradually flattens as it nears zero. Understanding how to sketch the exponential decay of the current is essential for visualizing electrical behaviors in time-dependent circuits.
Charge Accumulation and Time-Dependent Currents
When discussing charge accumulation, we're considering the total amount of charge that has been transferred over a period of time. In an electrical circuit, charge accumulation often occurs in components like capacitors. The current flowing into a device, such as a capacitor, leads to the accumulation of charge on its plates.

The relationship between current and charge is fundamental in electromagnetism: the current is the rate of charge flow. If we want to determine the total charge accumulated over a time period \(t_0\), given a time-varying current, we must integrate the current with respect to time from the start point to time \(t_0\).

The accumulated charge increases over time, and this increase will typically start faster and then slow down, approaching a maximum limit. This behavior is a result of the decaying current—fewer charges flow per unit time as the current diminishes. This concept is crucial in the design and operation of electronic timing circuits, such as those used in oscillators and timers.
Integral of Current to Find Accumulated Charge
The integral of current is a mathematical tool utilized to calculate the total charge transferred over a specific period. Importantly, knowing the current at any given moment is insufficient to find the total charge—it's necessary to consider the duration and the varying rate over that duration.

To perform this calculation, one integrates the current function with respect to time. If the current is constant, the integral simplifies to the product of current and time. However, in the case of time-varying currents, integration allows for the summation of infinitely many infinitesimal amounts of charge that flow for each moment.

The process starts by setting up the integral of the function representing the current over the specified time interval. After evaluating the integral, the result will provide the net charge transferred into or out of a component in the circuit over that interval. The step-by-step solution provided earlier depicted this integral approach for an exponentially decaying current. Students should remember that integrating current over time gives them the value of charge accumulated, a key concept in analyzing dynamic circuits.

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Most popular questions from this chapter

The current of electrons in a certain cathode-ray tube forms a thin cylindrical beam of constant radius \(a\) in which the charges move parallel to the axis. The current density in a cross-section is found not to be constant but to vary with the distance \(r\) from the axis of the beam according to \(J=J_{0}(1-r / a)\). What is the total current in the beam, assuming that it can be treated as the flow of a continuous distribution of charge? \(J_{0}\) is clearly the current density on the axis at \(r=0\). What is the mean current density in any cross-section in terms of \(J_{0}\) only?

The charge on a conductor at time \(t\) is given by \(Q=Q_{0}\left(1-e^{-i / T}\right)\) where \(Q_{0}\) and \(T\) are constants. Find the current \(I\) at time \(t\) and sketch the variations of \(Q\) and \(I\) with \(t\). If the charge were given instead by \(Q=Q_{0} e^{-U T}\) how would the current differ from that in the first case?

If it were possible to charge a conductor weighing \(10 \mathrm{~g}\) with \(1 \mathrm{C}\), what change in mass would occur on discharge to earth? How does this compare with that detectable by a chernical balance? (As to the feasibility of a charge of \(1 \mathrm{C}\), see Problems \(2.3\) and 4.6.).

Stow (1969) gives the total transfer of positive charge from the ionosphere to earth due to currents in fine weather as about \(90 \mathrm{C} \mathrm{km}^{-2}\) per year. What is the approximate fine weather current falling on \(1 \mathrm{~mm}^{2}\) of the earth? How many clectronic charges per second does this represent?

From the definition of the mole, show that the molar mass of any element in \(\mathrm{kg}\) is \(10^{-3}\) times its atomic mass in \(\mathrm{u}\), i.c. that the mass of 1 mol in \(\mathrm{g}\) is equal to its atomic mass in \(u\). The A vogadro constant \(N_{A}\) is the number of elementary units of any substance in 1 mol. Show that if \(m_{y}\) is the mass of \(1 \mathrm{u}\) in \(\mathrm{kg}\), then \(N_{A}=1 / 10^{3} m_{v}\) (For convenience, we denote \(10^{3} N_{A}\) in this book by \(N_{A}^{\prime}\) ) Show that a substance with relative atomic or molecular mass \(M\) and density \(\rho\) has a number of atoms or molecules per unit volume equal to \(N_{A} \rho / M\).
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