Question

Capacitance of a parallel plate capacitor becomes \((4 / 3)\) times its original value if a dielectric slab of thickness \(t=d / 2\) is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is (A) 8 (B) 4 (C) 6 (D) 2

Short answer

The dielectric constant of the slab is found by analyzing the given information of the capacitance of a parallel plate capacitor and the dielectric slab's thickness. After solving the equations, we find that the dielectric constant of the slab is 8. Therefore, the correct answer is (A) 8.

Step-by-step solution

Understand Capacitance and Dielectric Constant

Capacitance (C) of a parallel plate capacitor depends on the surface area (A) of the plates, the distance (d) between them, and the dielectric constant (k) of the material between the plates. In vacuum or air, k = 1, and the capacitance formula is: C = ε_0 * (A / d), where ε_0 is the vacuum permittivity. When a dielectric slab is introduced between the plates, the capacitance becomes: C' = k * ε_0 * (A / d).

Introduce a dielectric slab with thickness t

If we insert a dielectric slab of thickness t = d/2, the separation between the plates can be considered as being made up of two regions: one with the dielectric slab of thickness t and the other being empty (air, dielectric constant k = 1) of thickness (d - t). In this case, the effective capacitance C' is given by the formula for capacitors connected in series: (1 / C') = (1 / C₁) + (1 / C₂), where C₁ is the capacitance of the dielectric-filled region, and C₂ is the capacitance of the air-filled region. From the given information, we can write C' = (4/3)C.

Calculate C₁ and C₂

C₁ = K * ε_0 * (A / t), as it is filled with a dielectric slab, and C₂ = ε_0 * (A / (d-t)), as it is filled with air (k=1). Now substitute the values t = d/2, and (d-t) = d/2 in the equations of C₁ and C₂: C₁ = K * ε_0 * (A / (d/2)), C₂ = ε_0 * (A / (d/2)).

Find the relation between C' and dielectric constant

Now, using (1 / C') = (1 / C₁) + (1 / C₂), we can write: (1 / ((4/3)C)) = (1 / (K * ε_0 * (A / (d/2)))) + (1 / (ε_0 * (A / (d/2)))). Now, use C = ε_0 * (A / d): 1 / (4/3)C = 1 / (K * 2/1C) + 1 / (2/1C), Let's simplify the equation to find K: 3 = (8 + 4K) / K, 3K = 8 + 4K, K = 8.

Choose the correct option

Now, from the options given, we found that the dielectric constant of the slab is: (A) 8. Therefore, the correct answer is (A) 8.

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor is one of the simplest types of capacitors, known for its straightforward construction and operation. It consists of two large, flat plates that are placed parallel to each other, separated by a small distance. The space between these plates can be filled with air or any other insulating material, known as a dielectric. The main function of the capacitor is to store electrical energy in the electric field created by the separated charges on the plates.
The capacitance, which is the ability of the capacitor to store charge, is determined by three main factors:

• The surface area of the plates (A)
• The distance between the plates (d)
• The dielectric material between the plates.

In a vacuum, the formula for the capacitance of a parallel plate capacitor is given by:\[ C = \varepsilon_0 \frac{A}{d} \]Here, \(\varepsilon_0\) is the vacuum permittivity, a constant that characterizes the ability of a vacuum to permit electric field lines. This formula helps in calculating the charge stored for any given voltage applied across the capacitor's plates.

Dielectric Constant

The dielectric constant, often symbolized by \( k \), is a measure of a material's ability to increase the capacitance of a capacitor compared to when the capacitor is in a vacuum. By introducing a dielectric material between the plates of a capacitor, the capacitance is enhanced, allowing the capacitor to store more charge for the same applied voltage.
When we put a dielectric material in a capacitor, the new capacitance \( C' \) can be calculated using the formula:\[ C' = k \cdot \varepsilon_0 \cdot \frac{A}{d} \]Here, \( k \) is the dielectric constant of the material, and it effectively scales the vacuum capacitance up by a factor of \( k \).
Generally, dielectric constants are greater than 1, reflecting their property of increasing capacitance. In real-world applications, choosing the right dielectric material with a suitable dielectric constant is crucial for optimizing the efficiency and capacity of the capacitor.

Series Capacitance Formula

When multiple capacitors are arranged in series, their combined effect is different from just adding their individual capacitances. In the context of a parallel plate capacitor with a dielectric, this series combination occurs when part of the space is filled with a dielectric and the rest is air.
The formula for capacitors in series is given by:\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} \]In our example, where a dielectric slab of thickness \( t = \frac{d}{2} \) is used, you consider the capacitor as having two parts. One part with the dielectric (thickness \( t \)), and another part with air (thickness \( d-t \)).
To find the total capacitance, you calculate the capacitance of each section separately and then combine them using the series capacitance formula. This approach effectively provides the equivalent capacitance experienced by the capacitor when a combination of air and dielectric is involved.

Vacuum Permittivity

Vacuum permittivity, also denoted by \( \varepsilon_0 \), is a fundamental physical constant describing how electric fields interact with the vacuum of free space. It is a measure of the resistance encountered when forming an electric field in a vacuum, considered to be about \(8.854 \times 10^{-12} \, \text{F/m} \).

• This constant is essential in calculating the capacitance of capacitors without a dielectric.
• It acts as a baseline for understanding how different materials affect capacitance when used as a dielectric.
• By comparing capacitance values with and without a dielectric, \( \varepsilon_0 \) helps measure the effectiveness of the material inserted between capacitor plates.

Understanding vacuum permittivity is crucial for gaining insights into how electric fields are shaped and controlled in various environments, ultimately influencing the design and optimization of capacitors in high-tech and everyday electrical devices.