Problem 65 - University Physics with Modern Physics

Question

A dielectric of permittivity 3.5 \(\times\) 10\(^{-11}\) F/m completely fills the volume between two capacitor plates. For t 7 0 the electric flux through the dielectric is (8.0 \(\times\) 10\(^3\) V \(\cdot\) m\(/s^3)t^3\). The dielectric is ideal and nonmagnetic; the conduction current in the dielectric is zero. At what time does the displacement current in the dielectric equal 21 \(\mu\)A ?

Step-by-Step Solution

Answer

The displacement current equals 21 µA at approximately 0.3125 ms.

Step 1: Understanding the Formula for Displacement Current

The displacement current in a dielectric is given by the time derivative of the electric displacement field, i.e., \( I_D = \frac{d \Phi_E}{dt} \), where \( \Phi_E \) is the electric flux. We have \( \Phi_E = (8.0 \times 10^3 \text{ V} \cdot \text{m/s}^3)t^3 \).

Step 2: Calculate the Derivative of Electric Flux

Compute the derivative of the electric flux with respect to time to find the displacement current. Start by differentiating \( \Phi_E(t) = (8.0 \times 10^3 \text{ V} \cdot \text{m/s}^3)t^3 \):\[ \frac{d \Phi_E}{dt} = \frac{d}{dt}[(8.0 \times 10^3)t^3] = 3 \times (8.0 \times 10^3)t^2 = 24.0 \times 10^3 t^2 \text{ V} \cdot \text{m/s}^3. \]

Step 3: Relate Displacement Current to Time

The displacement current \( I_D \) in the dielectric is given by the time derivative of the electric flux, and we equate this to the given displacement current of 21 \( \mu\)A or 21 \times 10^{-6} A. Thus:\[ 24.0 \times 10^3 t^2 = 21 \times 10^{-6}. \]

Step 4: Solve for Time

Solve the equation \( 24.0 \times 10^3 t^2 = 21 \times 10^{-6} \) for \( t \):1. Rearrange the equation: \[ t^2 = \frac{21 \times 10^{-6}}{24.0 \times 10^3} \] 2. Substitute values and simplify: \[ t^2 = \frac{21}{24.0 \times 10^9} \] 3. Solve for \( t \): \[ t = \sqrt{\frac{21}{24.0 \times 10^9}}. \] 4. Calculate \( t \) using a calculator. This results in:\[ t \approx 3.125 \times 10^{-4} \text{ seconds}. \]

Key Concepts

Understanding Electric Flux

Electric flux is a concept that helps us understand the flow of electric field lines through a given area. Imagine a net placed in the rain; the amount of rain flowing through the net represents the flux. Similarly, electric flux quantifies the number of electric field lines passing through a surface.

Mathematically, electric flux \( \Phi_E \) is given by the formula \( \Phi_E = \vec{E} \cdot \vec{A} \), where \( \vec{E} \) is the electric field strength and \( \vec{A} \) is the area vector perpendicular to the field lines. When the area is aligned with the field, electric flux is maximized.

In the context of capacitors, electric flux becomes crucial as it relates to the electric field between the plates, particularly when a dielectric material is inserted between them. This helps in calculating displacement current, which is based on changes in electric flux over time.

Role of Dielectric Materials

Dielectric materials are insulating substances that can be polarized by an electric field. Unlike conductors, dielectrics do not allow current to flow through them but instead, help in storing electrical energy. This is why dielectrics are often used between the plates of capacitors.

The dielectric constant, or permittivity, is a measure of a material's ability to store electric charge. For an ideal dielectric, like the one described in the exercise with a permittivity of 3.5 \( \times \) 10\(^{-11}\) F/m, the ability to hold charge is effectively increased when this material is placed between capacitor plates.

By inserting a dielectric, the overall electric field within the capacitor is reduced for the same charge on the plates, lowering the voltage necessary to maintain the charge. This is directly related to the concept of displacement current, which arises even in the absence of any conduction current in the dielectric material.

Importance of Time Derivative in Displacement Current

The time derivative is a powerful concept in calculus that provides the rate of change of a quantity with respect to time. In our exercise, the time derivative is crucial in calculating the displacement current, which is the rate of change of the electric flux through the dielectric.

The displacement current \( I_D \) is calculated using the derivative \( \frac{d \Phi_E}{dt} \), where \( \Phi_E \) is the electric flux. In simpler terms, it tells us how quickly the electric field is changing over time between the plates of a capacitor. This notion helps us understand that even when there is no actual charge flow in the dielectric, changing electric fields can still produce effects similar to electrical currents.

Understanding the role of the time derivative is essential for determining how fast these changes occur and for calculating quantities like displacement current, which, in this case, becomes equal to a real current of 21 \(\mu\)A at a specific time.