Question

A parallel plate capacitor has the space between its plates filled by two slabs of thickness \((\mathrm{d} / 2)\) each and dielectric constant \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) If \(\mathrm{d}\) is the plate separation of the capacitor, then capacity of the capacitor is .......... (A) \(\left[\left(2 \mathrm{~d} \in_{0}\right) / \mathrm{A}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]\) (B) \(\left[\left(2 \mathrm{~A} \in_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right) /\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)\right]\) (C) \(\left[\left(2 \mathrm{Ad} \epsilon_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]\) d] \(\left(K_{1}+K_{2}\right)\) (D) \(\left[\left(2 \mathrm{~A} \in_{0}\right) /\right.\)

Short answer

The capacity of the capacitor is given by (B) \(\left[\left(2 \mathrm{~A} \epsilon_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right) /\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)\right]\).

Step-by-step solution

Understanding the problem

The key to this problem is understanding that the two dielectric slabs share the capacitor space evenly. Given that each slab has a thickness of \(d / 2\), they totally cover up the space between the capacitor plates. Since the slabs have different dielectric constants, we need to compute the capacitance contribution of each slab separately, and then compute the overall capacitance.

Finding the capacitance of each slab

The capacitance for each capacitor slab can be computed using the formula for capacitance of a parallel plate capacitor, which is \(C = \frac{\kappa \epsilon_0 A}{d}\). Since the dielectric fills half the distance between the plates for each slab, the thickness \(d/2\) should be used in the formula. Therefore, the capacitance for the dielectrics with dielectric constants \(K_1\) and \(K_2\) would be \(C_1 = \frac{K_1 \epsilon_0 A}{d/2}\) and \(C_2 = \frac{K_2 \epsilon_0 A}{d/2}\).

Computing the total capacitance

The total capacitance is the sum of the capacitances from each dielectric slab. However, we have to remember that these capacitances are in series, not in parallel. In series circuit, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances. Hence, the total capacitance \(C_{total}\) is given by: \(C_{total} = \frac{1} {(1/C_1)+(1/C_2)}\) Plugging in the expressions obtained in Step 2 and simplifying, one can find that \(C_{total} = \frac{(K_1 K_2) / (K_1 + K_2)}{\epsilon_0 A / d}\). This corresponds to answer choice (B).

Key concepts

Dielectric Constant

When discussing capacitors, particularly in the context of a parallel plate capacitor, the dielectric constant ( κ ) plays a pivotal role. A dielectric constant, also known as relative permittivity, is a measure of a substance's ability to store electrical energy in an electric field. In simple terms, it shows how well a material can get polarized under the application of an electric field.
The dielectric constant is always greater than or equal to 1. It affects the capacitance of a capacitor as it determines how much potential energy is stored. Different materials have different dielectric constants, which influences their suitability as a dielectric material in capacitors.

• A high dielectric constant means the material can store more charge at the same voltage.
• This concept is crucial in the design and function of capacitors, particularly when optimizing for certain capacitive properties.

Hence, understanding how different materials behave can help in designing capacitors that meet specific needs.

Capacitance Formula

The formula that describes the capacitance of a parallel plate capacitor is fundamental to understanding and calculating its value. This foundational formula is given by:\[ C = \frac{\kappa \epsilon_0 A}{d} \]Where:

• \(C\) is the capacitance.
• \(\kappa\) is the dielectric constant of the material between the plates.
• \(\epsilon_0\) is the permittivity of free space, a constant value.
• \(A\) is the area of one of the plates.
• \(d\) is the distance between the plates.

This equation illustrates that capacitance increases with a larger plate area or a smaller distance between plates. When a dielectric fills the space between the plates, the dielectric constant enhances the overall capacitance by increasing the ability to store electrical energy. This formula is central to the analysis and design of capacitors in various electrical circuits.

Series Capacitance

In the world of electrostatics, understanding how series capacitance operates is crucial. When capacitors are placed in series, the total or equivalent capacitance is less than any individual capacitance in the series. This concept can seem counterintuitive because we often associate combining things with increasing their cumulative effect. However, in series, the inverse becomes true.
The formula to find the total capacitance in a series configuration is:\[ C_{total} = \frac{1} {(1/C_1)+(1/C_2)+...+(1/C_n)} \]This formula indicates that the total capacitance decreases as more capacitors are added in series. Each added capacitor demands a share of the total voltage, spreading the potential difference across multiple stages. In the case of the parallel plate capacitor filled with two dielectric slabs, considering each slab as a capacitor allows the application of this principle.

• The inverse addition shows how individual contributions are combined into an overall smaller effect.
• This method highlights the cumulative limitations in charge storage when arranging capacitors in series.

Understanding series arrangements is a significant step toward grasping broader concepts in electrical engineering and electrostatics.

Electrostatics

Electrostatics encompasses the study of stationary charges or electric charges at rest. A crucial aspect of electrostatics is the interaction between electric fields and materials. When a parallel plate capacitor is involved, the electrostatic principles allow us to predict how charges will behave.
In a capacitor:

• An electric field is established between the plates when connected to a power source, causing the plates to accumulate electric charge.
• The direction of the electric field is from the positive to the negative plate.

Dielectric materials influence this field. They reduce the electric field strength inside the capacitor by getting polarized, which effectively increases the capacitor's capacitance. Electrostatic principles govern how these fields change and adjust depending on various factors, such as plate separation, the nature of the dielectric material, and the potential difference across the plates. A thorough grasp of electrostatics allows engineers and scientists to design more efficient electrical devices, ranging from capacitors to more complex systems, by predicting and controlling electric field behavior.