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?

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: I_{disp}=C⁢dVdt, where C is the capacitance and V is the voltage across the capacitor. Step 2: Substitute the given displacement current expression I_{disp}=10⁢A⁢exp(-t/2.0⁢μs) into the formula for displacement current. This gives: 10⁢exp(-t/2.0⁢μs)=1.0⁢μF⁢dVdt. Step 3: Rearrange the equation to isolate dVdt. This gives: dVdt=10⁢exp(-t/2.0⁢μs)1.0⁢μF. Step 4: Integrate both sides with respect to time to find the voltage V. The integral of the displacement current expression is: 10⁢exp(-t/2.0⁢μs)dt. Evaluate this integral to find the voltage as a function of time. Step 5: Determine the initial voltage (ΔV_{C})_{0} by evaluating the voltage expression at t=0. This involves substituting t=0 into the integrated voltage equation.

Ch 31: Electromagnetic Fields and Waves

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 31: Electromagnetic Fields and Waves

Problema 40

Capitolo 31, Problema 40

A 1.0 μF capacitor is discharged, starting at t = 0 s.The displacement current between the plates is $I_{disp} = (10 A) \exp (- \frac{t}{2.0} μ s)$ . What was the capacitor’s initial voltage (ΔV_C)₀?

Guida verificata passo dopo passo

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:

I

_disp=C⁢dVdt, where

C

is the capacitance and

V

is the voltage across the capacitor.

Step 2: Substitute the given displacement current expression

I

_disp=10⁢A⁢exp(-t/2.0⁢μs) into the formula for displacement current. This gives:

10 \exp (- t / 2.0 μs) = 1.0 μF \frac{d V}{dt}

Step 3: Rearrange the equation to isolate

\frac{d V}{dt}

. This gives:

\frac{d V}{dt} = \frac{10}{} 1.0 μF

Step 4: Integrate both sides with respect to time to find the voltage

V

. The integral of the displacement current expression is:

10 \exp (- t / 2.0 μs) d t

. Evaluate this integral to find the voltage as a function of time.

Step 5: Determine the initial voltage

(ΔV

_C)₀ by evaluating the voltage expression at

t = 0

. This involves substituting

t = 0

into the integrated voltage equation.

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Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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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Displacement vs. Distance

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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Amplitude Decay in an LRC Circuit

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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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Concetti chiave

Capacitance

Displacement Current

Exponential Decay

A 1.0 μF capacitor is discharged, starting at t = 0 s.The displacement current between the plates is $I_{disp} = (10 A) \exp (- \frac{t}{2.0} μ s)$ . What was the capacitor’s initial voltage (ΔV_C)₀?