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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 40

A 1.0 μF capacitor is discharged, starting at t = 0 s.The displacement current between the plates is Idisp=(10 A)exp(t2.0 μs)I_{\(\text{disp}\)}=(10\(\text{ A}\))\(\exp\]\left\)(-\(\frac{t}{2.0\text{ }\)}\(\mu\[\text{s}\]\right\)). What was the capacitor’s initial voltage (ΔVC)₀?

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Step 1: Understand the relationship between displacement current and the rate of change of electric field in a capacitor. The displacement current is given by the formula: Idisp=CdVdt, where C is the capacitance and V is the voltage across the capacitor.
Step 2: Substitute the given displacement current expression Idisp=10Aexp(-t/2.0μs) into the formula for displacement current. This gives: 10exp(-t/2.0μs)=1.0μFdVdt.
Step 3: Rearrange the equation to isolate dVdt. This gives: dVdt=10exp(-t/2.0μs)1.0μF.
Step 4: Integrate both sides with respect to time to find the voltage V. The integral of the displacement current expression is: 10exp(-t/2.0μs)dt. Evaluate this integral to find the voltage as a function of time.
Step 5: Determine the initial voltage (ΔVC)0 by evaluating the voltage expression at t=0. This involves substituting t=0 into the integrated voltage equation.

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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 capacitor to store charge per unit voltage. It is defined as C = Q/V, where C is capacitance in farads, Q is the charge in coulombs, and V is the voltage across the capacitor. In this case, the capacitor has a capacitance of 1.0 μF, which indicates how much charge it can hold at a given voltage.
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Capacitors & Capacitance (Intro)

Displacement Current

Displacement current is a concept introduced by James Clerk Maxwell to account for changing electric fields in capacitors. It is defined as Iₑₓₜ = ε₀(dΦ_E/dt), where ε₀ is the permittivity of free space and dΦ_E/dt is the rate of change of the electric field. In this problem, the displacement current is given as a function of time, indicating how the current changes as the capacitor discharges.
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Exponential Decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In the context of the displacement current, the equation Iₔᵢₛₚ = (10 A)exp(−t/2.0 μs) shows that the current decreases exponentially over time, which is characteristic of discharging capacitors. This behavior is crucial for determining the initial voltage across the capacitor.
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Related Practice
Textbook Question

At one instant, the electric and magnetic fields at one point of an electromagnetic wave are E=(200i^+300j^50k^) V/m\(\mathbf{E}\) = (200 \(\hat{\mathbf{i}\)} + 300 \(\hat{\mathbf{j}\)} - 50 \(\hat{\mathbf{k}\)}) \(\text{ V/m}\) and B=B0(7.3i^7.3j^+ak^ μT\(\mathbf{B}\)=B_0(7.3\(\hat{\mathbf{i}\)}-7.3\(\hat{\mathbf{j}\)}+a\(\hat{\mathbf{k}\)}\(\text{ }\]\mu\) T. What is the Poynting vector at this time and position?

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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

What is the total energy density in an electromagnetic wave of intensity 1000 W/m2?

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

At one instant, the electric and magnetic fields at one point of an electromagnetic wave are E=(200i^+300j^50k^) V/m\(\overrightarrow{E}\)=(200\(\hat{i}\)+300\(\hat{j}\)-50\(\hat{k}\))\(\text{ V/m}\) and B=B0(7.3i^7.3j^+ak^) μT\(\overrightarrow{B}\)=B_0(7.3\(\hat{i}\)-7.3\(\hat{j}\)+a\(\hat{k}\))\(\text{ }\[\mu\]\text{T}\). What are the values of aa and B0B_0?

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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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