Problem 60 - University Physics with Modern Physics

Question

A \(14.9-\mu F\) capacitor, a \(24.3-\mathrm{k} \Omega\) resistor, a switch, and a 25.-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 \(1.00 \mathrm{~cm}^{2}\) .

Step-by-Step Solution

Answer

Assume the area of the plates is \(1.00 cm^2\). Answer: To find the rate of change of the electric field between the plates of the capacitor at \(t=0.3621s\), follow these steps: 1. Convert all given values to standard units. 2. Find the voltage across the resistor using the RC circuit charging formula. 3. Calculate the voltage across the capacitor by subtracting the voltage across the resistor from the battery voltage. 4. Calculate the charge on the capacitor using the formula \(Q=CV_C\). 5. Calculate the electric field between the capacitor plates using the formula \(E(t)=\frac{Q}{A\epsilon_0}\), where \(\epsilon_0\) is the vacuum permittivity. 6. Differentiate the electric field equation concerning time and replace the voltage with the expression calculated before. Following these steps, the rate of change of the electric field at \(t = 0.3621s\) can then be determined.

Step 1: Convert all given values to standard units

Capacity of the capacitor: \(14.9 \mu F = 14.9 \times 10^{-6}F\) Resistance of the resistor: \(24.3 kΩ = 24.3 \times 10^3 Ω\) Area of the plates: \(1.00 cm^2 = 1.00 \times 10^{-4}m^2\)

Step 2: Find the voltage across the resistor

Using the RC circuit charging formula: \(V_R(t) = V_b * (1 - e^{-\frac{t}{RC}})\) \(V_R(0.3621s) = 25.0V * (1 - e^{-\frac{0.3621}{(14.9 \times 10^{-6})(24.3 \times 10^3)}})\)

Step 3: Calculate the voltage across the capacitor

At any time t, \(V_C(t) = V_b - V_R(t)\) So, at \(t = 0.3621s\), \(V_C(0.3621s) = 25 - V_R(0.3621s)\) Calculate the value of \(V_C(0.3621s)\).

Step 4: Calculate the charge on the capacitor

Using the formula: \(Q = CV_C\) Charge on the capacitor at \(t = 0.3621s\): \(Q(0.3621s) = (14.9 \times 10^{-6}) * V_C(0.3621s)\)

Step 5: Calculate the electric field between the capacitor plates

Use the formula: \(E(t) = \frac{V_C(t)}{d}\), where d is the distance between the capacitor plates. Since we weren't given the distance between the plates in the problem and we have the area of the plates (\(1.00 \times 10^{-4}m^2\)), we can use the relationship: \(E(t) = \frac{Q}{A \epsilon_0}\) Here, ε₀ is the vacuum permittivity, which is a constant with a value of approximately \(8.85 × 10^{-12} F/m\). \(E(0.3621s) = \frac{Q(0.3621s)}{(1.00 \times 10^{-4}) (8.85 × 10^{-12})}\)

Step 6: Calculate the rate of change of the electric field

Differentiate the electric field equation with respect to time and replace the charge with the expression obtained before: \(\frac{dE}{dt} = \frac{d}{dt}\left(\frac{Q(t)}{(1.00 \times 10^{-4}) (8.85 × 10^{-12})}\right)\) \(\frac{dE}{dt} (0.3621s) = \frac{dQ/dt(0.3621s)}{(1.00 \times 10^{-4}) (8.85 × 10^{-12})}\) Where \(\frac{dQ}{dt}\) can be found using the initial charge equation: \(Q(t)=CV_C(t)\) \(\frac{dE}{dt}(0.3621s) = \frac{d(C*V_C(t))/dt}{(1.00 \times 10^{-4}) (8.85 × 10^{-12})}\) Now, find the value of \(\frac{dE}{dt}(0.3621s)\). This is the rate of change of the electric field between the plates of the capacitor at \(t=0.3621s\).

Key Concepts

RC Circuit Charging

Understanding the principles of an RC circuit charging is fundamental to analyzing how electrical circuits work, particularly when involving resistors (R) and capacitors (C). When you close the switch in a circuit consisting of a resistor, capacitor, and a voltage source, the capacitor doesn't charge instantaneously. Instead, it undergoes a fascinating process whereby its voltage gradually increases and approaches the battery voltage with time.

As current flows through the resistor, a voltage drop occurs across it (following Ohm's Law), which thereby reduces the initial current over time. Consequently, the charging rate of the capacitor is not uniform; it follows an exponential function. Initially, the charging happens rapidly but slows down as the capacitor's voltage equals the battery's. This important concept of RC circuit charging helps us understand the dynamic relationship of current, voltage, and time as they influence one another within the circuit.

Capacitor Charge and Electric Field

When electric charge accumulates on a capacitor's plates, it creates an electric field in the region between them. This field is a vector quantity pointing from the positive to the negative plate and its strength, or magnitude, is proportional to the charge (Q) on the plates. We use the concept of an electric field (E) to describe the influence that a charged object exerts on other charges in the vicinity without making contact.

Mathematically, the electric field due to a capacitor can be determined by dividing the voltage across the capacitor (V) by the distance between the capacitor plates (d), typically given by the equation:
\(E = \frac{V_C}{d}\)
The electric field can also be expressed in terms of the charge on the capacitor and the area of the plates (A), using the constant of vacuum permittivity \(\rho_0\). This relationship is crucial as it lets us work out the electric field's strength at any given point in time for a charging or discharging capacitor.

Exponential Decay in Circuits

An exponential decay pattern is commonly observed in the processes of charging and discharging in RC circuits. It shows how a quantity decreases at a rate proportional to its current value, which is exactly what happens to the voltage across a charging or discharging capacitor in an RC circuit.

Through the lens of mathematics, this behavior is described by the equation:
\(V(t) = V_0 e^{-\frac{t}{\tau}}\),
where \(V(t)\) is the voltage at time t, \(V_0\) is the initial voltage, \(e\) is the base of the natural logarithm, t is time, and \(\tau\) is the time constant of the circuit (the product of R and C). This concept of exponential decay is invaluable since it provides a predictive tool for determining not just when a capacitor will be fully charged but also how it discharges over time, and thus, the rate of change of various quantities, including voltage and electric field, over time.

Differential Calculus in Physics

The field of differential calculus is a powerful tool employed in physics to describe how physical quantities change over time. When we look at the rate of change of the electric field in a capacitor, we are delving into a real-world application of differential calculus. To ascertain how quickly the electric field between the plates of a capacitor is changing at any point in time, we differentiate the expression that relates electric field strength to charge and time.

Here, the rate of change of the electric field (\(\frac{dE}{dt}\)) is conceptually the same as finding the derivative of the electric field with respect to time, which in this case involves applying the chain rule since the electric field depends on the charge Q, which itself changes over time as the capacitor charges. In essence, the tools of differential calculus not only allow us to analyze the instantaneous rates of change in physical systems but also enhance our understanding and control over these systems by predicting their behavior over time.