Question

A parallel plate air capacitor has a capacitance \(18 \mu \mathrm{F}\). If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes \(72 \mu \mathrm{F}\). The dielectric constant of the medium is (A) 4 (B) 12 (C) 9 (D) 2

Short answer

#Step 2: Modify the formula for the new capacitance# #tag_title#New Capacitance #tag_content#Let's call the new capacitance C₂, the area A₂, and the distance d₂. The distance is tripled, so d₂ = 3d₁. A dielectric medium is introduced, so the dielectric constant becomes K₂. Using the formula for capacitance, we have: C₂ = ε₀K₂A₂/d₂ Given C₂ = 72 µF #Step 3: Solve for the dielectric constant K₂# #tag_title#Dielectric Constant #tag_content#Since the area of the plates remains the same, we can write A₂ = A₁. Also, d₂ = 3d₁. Now, we can write the equation: \[ 72 = ε₀K₂\frac{A₁}{3d₁} \] Divide by the initial capacitance equation: \[ \frac{C₂}{C₁} = \frac{ε₀K₂\frac{A₁}{3d₁}}{ε₀\frac{A₁}{d₁}} \] Simplify the equation: \[ 4 = \frac{K₂}{3} \] Solve for K₂: \[ K₂ = 4 \times 3 = 12 \] The dielectric constant of the medium is 12. The correct answer is (B) 12.

Step-by-step solution

Initial Capacitance

We need to find the capacitance of the air capacitor before the changes were made. The formula for the capacitance of a parallel plate capacitor is C = ε₀KA/d Where C is the capacitance, ε₀ is the permittivity of free space, K is the dielectric constant, A is the area of the plates, and d is the distance between them. Since the initial capacitor is an air capacitor, K = 1. Let's call the initial capacitance C₁, the area A₁, and the distance d₁. So, C₁ = ε₀A₁/d₁ Given C₁ = 18 µF

Key concepts

Dielectric Medium

A dielectric medium is a non-conductive substance that can be placed between the plates of a capacitor to increase its capacitance. This material works by reducing the electric field within the capacitor, which allows the capacitor to store more charge for the same amount of voltage. Essentially, when a dielectric is introduced, it gets polarized and opposes the electric field caused by the charges on the plates. This opposition results in the ability to store additional charge, leading to increased capacitance.

Some typical examples of dielectric materials include glass, ceramic, plastic, and rubber. When comparing the performance of dielectric materials, their effectiveness is measured by a property known as the dielectric constant. In capacitive applications, using a dielectric medium can significantly affect the overall storage capacity and efficiency of the capacitor.

Parallel Plate Capacitor

A parallel plate capacitor is a simple electronic component used to store electrical energy. It consists of two conductive plates that are parallel to each other, separated by a certain distance, known as the dielectric distance or simply "d." These plates are often made of conductive material such as aluminum or copper and are placed a little apart to hold an electric field.

The capacitance of a parallel plate capacitor is given by the formula:

• \(C = \frac{\varepsilon A}{d}\)

Where \(C\) is the capacitance, \(\varepsilon\) is the permittivity of the medium between the plates, \(A\) is the area of one of the plates, and \(d\) is the distance between the plates. By altering these parameters, especially the distance \(d\) or area \(A\), the capacitance can be effectively adjusted. Parallel plate capacitors are fundamental components in electronics, used in a wide range of applications from simple circuits to complex electronic devices.

Dielectric Constant

The dielectric constant, also known as the relative permittivity, is a measure of a material's ability to store electrical energy in an electric field. It is symbolized by the letter \(K\) and indicates how much the dielectric medium can "boost" the capacitance of a capacitor compared to air or vacuum.

The relation for the capacitance in the presence of a dielectric is given as:

• \(C = \varepsilon_0 K \frac{A}{d}\)

Where \(\varepsilon_0\) is the permittivity of free space, \(K\) is the dielectric constant of the medium, \(A\) is the area of the plates, and \(d\) is the separation between them. A higher dielectric constant means that the material is more efficient at increasing the capacitance of a capacitor. For example, in the given exercise, the dielectric constant was found to be 4, indicating the medium made the capacitance four times that of the air-filled capacitor, assuming the same physical dimensions and absence of changes from the introduction of the dielectric.

Permittivity of Free Space

The permittivity of free space, denoted by \(\varepsilon_0\), is a fundamental physical constant that describes how much electric field (electrical force per unit charge) is generated in a vacuum. This constant is crucial in the calculation of electric forces and plays a key role in determining the capacitance of a capacitor in a vacuum or air.

The value of the permittivity of free space is approximately \(8.85 \times 10^{-12} \text{F/m}\) (farads per meter). This parameter is embedded in the formula for the capacitance of a parallel plate capacitor:

• \(C = \varepsilon_0 K \frac{A}{d}\)

Where \(C\) is the capacitance, \(K\) is the dielectric constant, \(A\) is the area of the plates, and \(d\) is the distance between them. The permittivity of free space serves as a baseline for calculating how a material placed between the capacitor plates affects its total capacitance. Even though \(\varepsilon_0\) is constant, its presence allows us to determine the effect of other materials as expressed through the dielectric constant.