Question

A parallel plate capacitor with \(C=0.050 \mu \mathrm{F}\) has a separation between its plates of \(d=50.0 \mu \mathrm{m} .\) The dielectric that fills the space between the plates has dielectric constant \(\kappa=2.5\) and resistivity \(\rho=4.0 \cdot 10^{12} \Omega \mathrm{m} .\) What is the time constant for this capacitor? (Hint: First calculate the area of the plates for the given \(C\) and \(\kappa\), and then determine the resistance of the dielectric between the plates.)

Short answer

The time constant of the capacitor is approximately 8500 seconds.

Step-by-step solution

Calculate the area of the plates

To calculate the area of the plates, we should first rearrange the formula for capacitance: \(A = \frac{Cd}{\kappa \epsilon_0}\). We are given \(C = 0.050 \mu \mathrm{F}\), \(d = 50.0 \mu \mathrm{m}\), and \(\kappa = 2.5\). We also know that the vacuum permittivity constant \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\). Plugging in these values, we get: $$A = \frac{(0.050 \times 10^{-6} \mathrm{F})(50.0 \times 10^{-6} \mathrm{m})}{2.5 (8.85 \times 10^{-12} \mathrm{F/m})} \approx 1.12 \times 10^{-3} \mathrm{m^2}$$

Calculate the resistance of the dielectric

Now, we need to calculate the resistance \(R\) of the dielectric using the formula \(R = \frac{\rho L}{A}\). We are given the resistivity \(\rho = 4.0 \times 10^{12} \Omega \mathrm{m}\) and the separation between the plates \(d = 50.0 \mu \mathrm{m} = L\). We have found the area \(A = 1.12 \times 10^{-3} \mathrm{m^2}\). Plugging in these values, we get: $$R = \frac{(4.0 \times 10^{12} \Omega \mathrm{m})(50.0 \times 10^{-6} \mathrm{m})}{1.12 \times 10^{-3} \mathrm{m^2}} \approx 1.7 \times 10^8 \Omega$$

Calculate the time constant

Finally, we can calculate the time constant \(\tau\) using the formula \(\tau = RC\). We have found \(R = 1.7 \times 10^8 \Omega\) and we were given \(C = 0.050 \mu \mathrm{F}\). Plugging in these values, we get: $$\tau = (1.7 \times 10^8 \Omega)(0.050 \times 10^{-6} \mathrm{F}) \approx 8500~\mathrm{s}$$ The time constant for this capacitor is approximately 8500 seconds.

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor is a fundamental electronic component that comprises two conductive plates separated by an insulating material, known as a dielectric. The plates are kept parallel to each other, creating a uniform electric field between them. This setup allows the capacitor to store electrical energy. The capacitance, which is the ability of the capacitor to store charge, depends on several factors:

• Area of the plates: Larger plates can store more charge.
• Distance between the plates: Closer plates increase the capacitance.
• Properties of the dielectric material placed between the plates.

A parallel plate capacitor is often used in circuits to smooth out fluctuations in electrical signals, filter out noise, store energy temporarily, or in various timing applications due to its ability to release charge at a controlled rate.

Dielectric Constant

The dielectric constant, often symbolized as \(\kappa\), is a measure of a material's ability to store electrical energy in an electric field. It is a dimensionless number that indicates the amount by which the dielectric material can increase the capacitance of a capacitor compared to a vacuum. A higher dielectric constant means that the material is more effective at storing charge.

When a dielectric is inserted between the plates of a capacitor,

• It reduces the effective electric field.
• This leads to an increase in the overall capacitance.

In practical terms, by using a dielectric with a higher dielectric constant, we can design capacitors with greater capacitance values without increasing the size of the plates or decreasing the distance between them. Dielectrics are crucial in designing efficient and miniaturized electronic components.

Resistance of Dielectric

The resistance of a dielectric material, calculated using the formula \(R = \frac{\rho L}{A}\), is a measure of how much the material resists the flow of electric current. Here, \(\rho\) is the resistivity of the material, \(L\) is the separation between the plates (also understood as the dielectric thickness), and \(A\) is the area of the plates.

Key points to consider:

• High resistivity ensures that the dielectric prevents current from leaking across the plates.
• Dielectric materials are used not only to increase capacitance but also to insulate against unwanted current flow.

An appropriate selection of the dielectric material ensures that the capacitor performs efficiently, storing charge without significant leakage, which would otherwise reduce its effectiveness in a circuit.

Capacitance Formula

The capacitance of a parallel plate capacitor can be calculated using the formula \(C = \frac{\kappa \epsilon_0 A}{d}\), where \(C\) is the capacitance, \(\epsilon_0\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the separation between the plates.

To design a capacitor with a specific capacitance value, one can:

• Increase the plate area to allow more charge storage.
• Improve the efficiency of the capacitor by selecting materials with a high dielectric constant.
• Reduce the distance between the plates, allowing stronger electric field influence.

Understanding this formula is essential when designing electronics, as capacitors are widely used in applications requiring precise charge management and timing characteristics. Efficient capacitance design can greatly influence overall device performance.