Problem 72 - University Physics with Modern Physics

Question

A capacitor has two parallel plates with area \(A\) separated by a distance \(d\) . The space between plates is filled with a material having dielectric constant \(K\) . The material is not a perfect insulator but has resistivity \(\rho\) . The capacitor is initially charged with charge of magnitude \(Q_{0}\) on each plate that gradually discharges by conduction through the dielectric. (a) Calculate the conduction current density \(j_{\mathrm{C}}(t)\) in the dielectric. (b) Show that at any instant the dis-placement current density in the diclectric is equal in magnitude to the oonduotion current density but opposite in direction, so the total current density is zero at every instant.

Step-by-Step Solution

Answer

The conduction current density is \( j_C(t) = \frac{Q(t)}{K\varepsilon_0 A \rho} \). The total current density is zero at every instant because the conduction and displacement current densities cancel each other out.

Step 1: Understanding the Parameters

The problem involves a capacitor with parallel plates area \( A \) and separation \( d \). The space between them is filled with a dielectric material with dielectric constant \( K \) and resistivity \( \rho \). The capacitor initially holds a charge \( Q_0 \), and it discharges over time.

Step 2: Ohm's Law for Conduction Current Density

The conduction current density \( j_{\mathrm{C}}(t) \) is given by Ohm's Law, \( j_{\mathrm{C}}(t) = \frac{1}{\rho} E(t) \), where \( E(t) \) is the electric field across the dielectric. Since \( V(t) = E(t) \cdot d \), we use the relation \( E(t) = \frac{V(t)}{d} \). Thus, \( j_{\mathrm{C}}(t) = \frac{V(t)}{\rho d} \).

Step 3: Express Voltage in Terms of Charge

The voltage across the plates is \( V(t) = \frac{Q(t)}{C} \) where \( C = \frac{K\varepsilon_0 A}{d} \) is the capacitance. Thus, \( V(t) = \frac{Q(t) d}{K\varepsilon_0 A} \).

Step 4: Relating Current Density to Charge

Substituting the expression for \( V(t) \) into the current density equation gives:\[ j_{\mathrm{C}}(t) = \frac{Q(t) d}{K\varepsilon_0 A \rho d} = \frac{Q(t)}{K\varepsilon_0 A \rho} \].

Step 5: Show Displacement Current Density Relationship

The displacement current density \( j_D(t) \) is given by \( j_D(t) = \varepsilon \frac{dE}{dt} \), where \( \varepsilon = K\varepsilon_0 \). Since \( j_D(t) = -j_C(t) \), we evaluate \( \frac{dE}{dt} \) to show this relation.

Step 6: Charge Decay Over Time

Assume \( Q(t) = Q_0 e^{-t/\tau} \) where \( \tau = \rho \varepsilon \). Then, \( \frac{dQ}{dt} = -\frac{Q_0}{\tau} e^{-t/\tau} \).

Step 7: Final Verification of Zero Total Current Density

Substitute \( \frac{dQ}{dt} \) into the displacement current density formula, and verify \( j_D(t) = -j_C(t) \). This shows that the total current density \( j_{\text{total}}(t) = j_C(t) + j_D(t) = 0 \), confirming the problem statement.

Key Concepts

Capacitor

A capacitor is an essential component in electronic circuits. It stores electrical energy in an electric field between two conductive plates separated by an insulating material. These plates can store a charge when connected to a power source. The amount of charge they can store is determined by the capacitance, which depends on factors such as the plate area, the separation distance, and the dielectric material between the plates.
The capacitance of a parallel plate capacitor, for example, is given by the formula:

• \( C = \frac{K\varepsilon_0 A}{d} \)

where \( C \) is the capacitance, \( K \) is the dielectric constant, \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of one of the plates, and \( d \) is the separation between the plates.
Capacitors are used to filter signals, store energy, and stabilize voltage and power flow in various applications.

Dielectric

The dielectric material between the plates of a capacitor plays a crucial role in its function. It is an insulating material that increases the capacitor's ability to store energy by reducing the electric field strength needed to hold a given amount of charge.
Dielectric materials have several key properties:

• Dielectric Constant (\( K \)): This quantifies how much electric charge the material can hold relative to a vacuum. A higher dielectric constant means more charge storage capacity.
• Resistivity (\( \rho \)): This measures how strongly a material opposes the flow of electric current. Even in dielectrics, some tiny current may flow due to imperfections.

The dielectric constant and resistivity affect how a capacitor charges and discharges over time, influencing its behavior in circuits.
In the exercise we discussed, the dielectric not only stores charge but also allows some conduction of electricity, leading to a discharge of the capacitor.

Current Density

Current density is a measure of the electric current per unit area of a cross-section. In capacitors, especially those with dielectric materials, both conduction and displacement current densities are essential concepts to understand.

• Conduction Current Density (\( j_C(t) \)): This occurs due to the flow of charge through a dielectric and is described by Ohm's law: \( j_C(t) = \frac{1}{\rho} E(t) \), where \( E(t) \) is the electric field, and \( \rho \) is the material's resistivity.
• Displacement Current Density (\( j_D(t) \)): This is an apparent current that appears when the electric field changes, even when no real charge moves. It is crucial in understanding how capacitors work in alternating current circuits.

The exercise demonstrates that these current densities can cancel each other, resulting in zero total current density at any instant, which means the net current within the dielectric is balanced.

Electric Field

An electric field is a region around a charged object where other charges feel a force. In capacitors, the electric field is the force exerted by the potential difference between the plates, helping to store energy.
The electric field related to a capacitor is given by

• \( E = \frac{V}{d} \)

where \( V \) is the voltage, and \( d \) is the separation between the plates.
The electric field's strength can influence the flow of current and affect how energy is stored and released in a capacitor. In the case of the dielectric, the electric field can also lead to conduction and displacement currents. The balance of these currents is vital to ensure the proper functioning of capacitors in circuits.