The capacitor in an RC circuit is discharged with a time constant of 10 ms. At what time after the discharge begins are (a) the charge on the capacitor reduced to half its initial value and (b) the energy stored in the capacitor reduced to half its initial value?
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Problem 70b
Textbook Question
Two 10-cm-diameter metal plates 1.0 cm apart are charged to ±12.5 nC. They are suddenly connected together by a 0.224-mm-diameter copper wire stretched taut from the center of one plate to the center of the other. Does the current increase with time, decrease with time, or remain steady? Explain.

1
Step 1: Understand the physical setup of the problem. Two metal plates are charged to ±12.5 nC and connected by a copper wire. The plates act as a capacitor, and the wire provides a path for the charges to flow, creating a current. The question asks about the behavior of this current over time.
Step 2: Recall the concept of a discharging capacitor. When a capacitor is connected through a conductor, the charges on the plates flow to neutralize each other. The current is initially high because the potential difference between the plates is large, but it decreases over time as the charge on the plates diminishes.
Step 3: Use the formula for the current in a discharging capacitor: \( I(t) = I_0 e^{-t/RC} \), where \( I_0 \) is the initial current, \( R \) is the resistance of the wire, \( C \) is the capacitance of the plates, and \( t \) is time. This equation shows that the current decreases exponentially with time.
Step 4: Consider the physical properties of the wire and plates. The resistance \( R \) of the copper wire can be calculated using \( R = \frac{\rho L}{A} \), where \( \rho \) is the resistivity of copper, \( L \) is the length of the wire, and \( A \) is its cross-sectional area. The capacitance \( C \) of the plates can be calculated using \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between them.
Step 5: Conclude that the current decreases with time. This is because the charge on the plates diminishes as the capacitor discharges, reducing the potential difference and, consequently, the current flowing through the wire. The exponential decay behavior is characteristic of a discharging capacitor.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Capacitance
Capacitance is the ability of a system to store electric charge per unit voltage. In this scenario, the two metal plates form a capacitor, where the capacitance depends on their area and the distance between them. When charged, the plates create an electric field, and the stored energy can influence the current flow when they are connected.
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Capacitors & Capacitance (Intro)
Current Flow in Circuits
Current is the flow of electric charge through a conductor, measured in amperes. When the charged plates are connected by a wire, the potential difference between them drives the current. The behavior of the current over time is influenced by the capacitance of the plates and the resistance of the wire, which can cause the current to change as the system reaches equilibrium.
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Flow Continuity
RC Time Constant
The RC time constant is a measure of the time it takes for the current to charge or discharge in a resistor-capacitor circuit. It is calculated as the product of resistance (R) and capacitance (C). In this case, as the plates discharge through the wire, the current will initially be high and then decrease over time, following an exponential decay governed by the RC time constant.
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Phase Constant of a Wave Function
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