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Ch 31: Electromagnetic Fields and Waves
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 31, Problem 34b

A simple series circuit consists of a 150 Ω resistor, a 25 V battery, a switch, and a 2.5 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t = 0 s. Find the electric flux and the displacement current at t = 0.50 ns.

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Step 1: Understand the problem. The circuit consists of a resistor, a capacitor, and a battery in series. At t = 0, the capacitor is uncharged, and the switch is closed. We are tasked with finding the electric flux and the displacement current at t = 0.50 ns. The displacement current is related to the rate of change of the electric field in the capacitor, and the electric flux is related to the electric field between the plates.
Step 2: Calculate the charge on the capacitor at t = 0.50 ns. The charge on the capacitor as a function of time in an RC circuit is given by the equation: q=Q(1-e-), where Q is the maximum charge (Q = C × V), and the time constant τ = RC. Substitute the given values: R = 150 Ω, C = 2.5 pF, and V = 25 V.
Step 3: Determine the electric field between the plates of the capacitor. The electric field is related to the charge on the capacitor by the formula: E=q/A, where A is the area of the plates. The area can be calculated using the capacitance formula for a parallel-plate capacitor: C=εA/d, where ε is the permittivity of free space and d is the separation between the plates.
Step 4: Calculate the electric flux. The electric flux is given by the formula: Φ=EA. Use the electric field calculated in Step 3 and the area of the plates to find the flux.
Step 5: Calculate the displacement current. The displacement current is related to the rate of change of the electric flux by the formula: Id=ε/dt. Differentiate the electric flux with respect to time to find the displacement current at t = 0.50 ns.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It is calculated as the product of the electric field strength and the area perpendicular to the field. In the context of capacitors, electric flux can be related to the charge stored in the capacitor and the area of its plates, providing insight into how electric fields interact with materials.
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Displacement Current

Displacement current is a concept introduced by James Clerk Maxwell to account for changing electric fields in situations where there is no actual current flow, such as in capacitors. It is defined as the rate of change of electric flux through a surface and is crucial for understanding how electric fields can produce magnetic fields, even in the absence of conduction current.
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Capacitance and Charging of Capacitors

Capacitance is the ability of a capacitor to store charge per unit voltage, defined as C = Q/V, where Q is the charge and V is the voltage across the capacitor. When a capacitor is connected to a voltage source, it begins to charge, and the current flowing into the capacitor decreases over time as it approaches its maximum charge. Understanding the charging process is essential for calculating the electric flux and displacement current at any given time.
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Related Practice
Textbook Question

In FIGURE P31.32, a circular loop of radius r travels with speed v along a charged wire having linear charge density λ. The wire is at rest in the laboratory frame, and it passes through the center of the loop. What electric and magnetic fields would an experimenter in the loop's frame calculate at distance r from the current of part c?

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Textbook Question

A proton is fired with a speed of 1.0×106 m/s through the parallel-plate capacitor shown in FIGURE P31.29. The capacitor's electric field is E =(1.0×105 V/m, down). How does an experimenter in the proton's frame explain that the proton experiences no force as the charged plates fly by?

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Textbook Question

A wire with conductivity σ carries current I. The current is increasing at the rate dI/dt. Evaluate the displacement current for a copper wire in which the current is increasing at 1.0×106 A/s.

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Textbook Question

An electron travels with v=(5.0×106i^)m/s\(\vec{v}\) = (5.0 \(\times\) 10^6\,\(\hat{i}\))\,\(\text{m/s}\) through a point in space where E=(2.0×105i^2.0×105j^)V/m\(\vec{E}\) = (2.0 \(\times\) 10^5\,\(\hat{i}\) - 2.0 \(\times\) 10^5\,\(\hat{j}\))\,\(\text{V/m}\) and B=0.10k^T\(\vec{B}\) = -0.10\,\(\hat{k}\)\,\(\text{T}\). What is the force on the electron?

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Textbook Question

A 10 A current is charging a 1.0-cm-diameter parallel-plate capacitor. What is the magnetic field strength at a point 2.0 mm radially from the center of the capacitor?

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Textbook Question

FIGURE P31.38 shows the electric field inside a cylinder of radius R=3.0R=3.0 mm. The field strength is increasing with time as E=1.0×108t2E=1.0\(\times\)10^8t^{2} V/m, where t is in s. The electric field outside the cylinder is always zero, and the field inside the cylinder was zero for t<0t<0. Find an expression for the electric flux ΦeΦ_e through the entire cylinder as a function of time.

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