Question

A \(14.9-\mu F\) capacitor, a \(24.3-\mathrm{k} \Omega\) resistor, a switch, and a \(25.0-\mathrm{V}\) battery are connected in series. What is the rate of change of the electric field between the plates of the capacitor at \(t=0.3621 \mathrm{~s}\) after the switch is closed? The area of the plates is \(1.00 \mathrm{~cm}^{2}\)

Short answer

Answer: The rate of change of the electric field between the plates of the capacitor at t=0.3621 seconds after the switch is closed is approximately 1.876 * 10^{11} V/m/s.

Step-by-step solution

Determine the time constant (τ) of the RC circuit

The time constant of an RC circuit (resistor-capacitor circuit) is given by the formula: τ = RC Where R is the resistance and C is the capacitance. Plug in the values given in the exercise: τ = (24.3 * 10^3 Ω) * (14.9 * 10^{-6} F ) = 0.36207 seconds

Find the charge on the capacitor as a function of time (Q(t))

To find the charge as a function of time, we use the formula: Q(t) = Q₀ (1 - e^{(-t/τ)}) Where Q₀ is the maximum charge on the capacitor and is given by: Q₀ = C * V Plug in the values given in the exercise: Q₀ = (14.9 * 10^{-6} F) * (25.0 V) = 3.725 * 10^{-4} C

Calculate the rate of change of Q(t) with respect to time (dQ/dt)

To find the rate of change of Q(t), we need to find the derivative of Q(t) with respect to time (t): dQ/dt = d(Q₀ (1 - e^{(-t/τ)}))/dt Using the chain rule, we can find the derivative: dQ/dt = Q₀ * (e^{(-t/τ)}) * (1/τ) Plug in the values from Step 1 and 2 (τ and Q₀) and the given time (t=0.3621s): dQ/dt = (3.725 * 10^{-4} C) * (e^{(-0.3621/0.36207)}) * (1/0.36207) ≈ 1.662 * 10^{-4} A The rate of change of charge on the capacitor at t=0.3621 seconds is approximately 1.662 * 10^{-4} A.

Calculate the electric field E between the plates

The electric field E between the plates of the capacitor can be found using the formula: E = Q(t) / (ε₀ * A) Where ε₀ is the vacuum permittivity, which is approximately 8.854 * 10^{-12} F/m, and A is the area of the plates. We will first calculate the electric field as a function of time: E(t) = Q(t) / (ε₀ * A) Plug in the values of ε₀ and the given area A: E(t) = Q(t) / (8.854 * 10^{-12} F/m * 1.00 * 10^{-4} m²)

Calculate the rate of change of the electric field

Now we need to find the rate of change of the electric field with respect to time (dE/dt). Since Q(t) and A are the only variables, we can rewrite the formula for E(t) as: dE/dt = (dQ/dt) / (ε₀ * A) Plugging in the values of dQ/dt, ε₀, and A, we get: dE/dt = (1.662 * 10^{-4} A) / (8.854 * 10^{-12} F/m * 1.00 * 10^{-4} m²) dE/dt ≈ 1.876 * 10^{11} V/m/s The rate of change of the electric field between the plates of the capacitor at t=0.3621 seconds after the switch is closed is approximately 1.876 * 10^{11} V/m/s.

Key concepts

Understanding Capacitor Discharge

A fundamental concept in electrical circuits is how a capacitor discharges over time. A capacitor, which stores electric energy when connected to a power source like a battery, begins to release this energy once the power source is disconnected or a circuit is completed through a resistor. The discharge process is not instantaneous but follows an exponential decay.

Specifically, the voltage across the capacitor's plates and the charge it holds drop following a formula generally expressed as: \[\begin{equation}V(t) = V_0 e^{-t/\tau}\text{ and }Q(t) = Q_0 e^{-t/\tau}\text{, respectively,}\text{ where }\tau = RC\text{ is the time constant.}\text{ Here, }V_0\text{ and }Q_0\text{ represent the initial voltage and charge, }R\text{ is the resistance, }C\text{ is the capacitance, and }t\text{ is the time elapsed since the discharge began.}
For the exercise given, recognizing the exponential nature of capacitor discharge helps explain why the rate of change of charge, and consequently the rate of change of the electric field between the plates, also follows an exponential decay. This behavior underpins the observed pattern in which the rate of change is highest initially and decreases over time.

Electric Field Change in a Capacitor

When studying capacitors, it's important to realize that the electric field (\(E\)) between the plates is directly related to the charge on the capacitor. For a parallel plate capacitor, the electric field can be described by the equation: \[\begin{equation} E = \frac{Q}{\varepsilon_0 A} \text{, where } Q\text{ is the charge, }\varepsilon_0\text{ is the vacuum permittivity, and }A\text{ is the area of the plates.} \text{ As the capacitor discharges, the charge decreases, which causes the electric field to change.} The rate at which this field changes is of particular interest in various applications, including signal processing and the operation of electronic devices. To quantify this rate, one must look at the rate of charge change (\(dQ/dt\)) and how it affects the electric field over time. In more complex systems, electric field changes can induce currents, influence signal propagation, and affect how energy is stored and released.

In the exercise example, once the rate of change of charge is calculated, it can be directly related to the rate of change of the electric field since the area of the plates and vacuum permittivity are constants. The resulting value offers insight into how rapidly the electric field strength is diminishing between the plates as the capacitor discharges.

Analysis of Resistor-Capacitor (RC) Circuits

Resistor-Capacitor (RC) circuits are foundational in understanding electrical circuit behavior, especially in terms of charge and discharge cycles. They are typically used to create delays or time-based operations within electronic devices. Analyzing an RC circuit involves understanding both the transient and steady-state behavior of the circuit.

During the transient state, immediately after the circuit is switched on or off, the current and voltage do not change instantaneously but vary over time according to exponential functions. This is where the time constant (\(\tau\)) becomes a crucial parameter, as it determines how quickly the circuit will reach a certain percentage of its final steady state value. The relationship between the capacitor's voltage, the charge on the capacitor, and the current through the resistor all follow this exponential trend governed by the time constant.

An understanding of calculus is essential for correctly analyzing the rates of change at any given time, which is why for exercises like the one provided, finding derivatives is a key step. Through such analysis, we can describe the circuit's behavior mathematically and predict how it will behave at any point in time after the switch is closed, supporting the design and troubleshooting of electronic systems.