Problem 109 - Master Resource Book in JEE Main Physics

Question

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness \(d_{1}\) and dielectric constant \(K_{1}\) and the other has thickness \(d_{2}\) and dielectric constant \(K_{2}\) as shown in figure. This arrangement can be thought as a dielectric slab of thickness \(d\left(=d_{1}+d_{2}\right)\) and effective dielectric constant \(K .\) The \(K\) is \(\quad\) [NCERT Exemplar] (a) \(\frac{K_{1} d_{1}+K_{2} d_{2}}{d_{1}+d_{2}}\) (b) \(\frac{K_{1} d_{1}+K_{2} d_{2}}{K_{1}+K_{2}}\) (c) \(\frac{K_{1} K_{2}\left(d_{1}+d_{2}\right)}{\left(K_{1} d_{1}+K_{2} d_{2}\right)}\) (d) \(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\)

Step-by-Step Solution

Answer

The correct answer is (c) \( \frac{K_1 K_2 (d_1 + d_2)}{K_1 d_2 + K_2 d_1} \).

Step 1: Understand the Problem Statement

You have a parallel plate capacitor with two dielectric materials in series. Each material has its own thickness and dielectric constant, and you aim to find the effective dielectric constant (\( K \)) for the combined thickness \( d = d_1 + d_2 \).

Step 2: Conceptualize the Scenario

In a capacitor with dielectrics in series, the overall effect can be considered as having a single equivalent dielectric slab. We assume that the electric displacement \( D \) remains constant and that dielectrics in series add inversely like capacitors in series.

Step 3: Expression for Capacitance with Dielectrics in Series

For capacitors in series, the reciprocal capacitance is the sum of the reciprocals of the individual capacitances: \[ \frac{1}{C} = \frac{d_1}{K_1 \varepsilon_0 A} + \frac{d_2}{K_2 \varepsilon_0 A} \]Where \( A \) is the area of the plates and \( \varepsilon_0 \) is the permittivity of free space.

Step 4: Effective Capacitance Expression

The effective capacitance \( C_{eff} \) for the series arrangement is: \[ \frac{1}{C_{eff}} = \frac{1}{C_1} + \frac{1}{C_2} \] Substitute for \( C_1 \) and \( C_2 \), \[ \frac{1}{C_{eff}} = \frac{d_1}{K_1 \varepsilon_0 A} + \frac{d_2}{K_2 \varepsilon_0 A} \] Simplified, \[ C_{eff} = \frac{K_{eff} \varepsilon_0 A}{d_1 + d_2} \] Where \( K_{eff} \) is the effective dielectric constant.

Step 5: Derive Effective Dielectric Constant

Set the calculated \( C_{eff} \) equal to the series capacitance equation: \[ \frac{K_{eff} \varepsilon_0 A}{d} = \frac{K_1 \varepsilon_0 A}{d_1} + \frac{K_2 \varepsilon_0 A}{d_2} \] Simplify: \[ \frac{1}{K_{eff}} = \frac{d_1}{K_1 (d_1 + d_2)} + \frac{d_2}{K_2 (d_1 + d_2)} \] Taking the reciprocal, we get: \[ K = \frac{K_1 K_2 (d_1 + d_2)}{K_1 d_2 + K_2 d_1} \]

Step 6: Match with Given Options

Compare with the given answer choices:- Option (a): \( \frac{K_1 d_1 + K_2 d_2}{d_1 + d_2} \)- Option (b): \( \frac{K_1 d_1 + K_2 d_2}{K_1 + K_2} \)- Option (c): \( \frac{K_1 K_2 (d_1 + d_2)}{K_1 d_2 + K_2 d_1} \)- Option (d): \( \frac{2 K_1 K_2}{K_1 + K_2} \)Our calculated expression matches option (c).

Key Concepts

Dielectric Constant

The dielectric constant, often represented as "K," is a crucial concept in understanding how materials influence electric fields. When you insert a dielectric material between capacitor plates, it affects the capacitor's ability to store charge by reducing the electric field. This reduction occurs because the dielectric material becomes polarized in the presence of the electric field, which lessens the field's strength inside the material.

- **Definition**: The dielectric constant is a measure of a material's ability to insulate electric charges from each other. It's expressed as the ratio of the material's permittivity to the permittivity of free space (\( \varepsilon_0 \)).- **Mathematical Representation**: It is calculated by the formula \( K = \frac{\varepsilon}{\varepsilon_0} \), where \( \varepsilon \) is the permittivity of the dielectric material.- **Role in Capacitors**: A higher dielectric constant means the material can store more electric energy per unit volume under a given electric field. Hence, it enhances the capacitance of a capacitor by \( K \times \varepsilon_0 \times \frac{A}{d} \), where \( A \) is the plate area and \( d \) is the separation between plates.

In this problem, we are interested in finding the effective dielectric constant \( K_eff \) when two dielectric blocks with different thicknesses and constants are placed in series.

Capacitors in Series

When capacitors are connected in series, the total capacitance is not simple to find directly. Unlike capacitors in parallel, where capacitance adds up, capacitors in series require a reciprocal relationship.

- **Rule for Series Connection**: For capacitors in series, the total or effective capacitance \( C_{eff} \) is calculated using the formula \[ \frac{1}{C_{eff}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n} \]- **Understanding the Concept**: In a series setup, the charge \( Q \) on each capacitor remains the same, while the voltage across each capacitor is different. Total voltage \( V \) is the sum of individual voltages.- **Application to Dielectrics**: When dielectric layers are in series within a capacitor, each layer acts as a separate capacitor. Thus, to find out how their dielectric constants contribute to the whole, the capacitance formula takes the aforementioned reciprocal form.- **Current Exercise**: The problem involves two dielectric materials. The overall effect on the capacitor is similar to finding effective capacitance from each dielectric slab which impacts how we calculate the effective dielectric constant.

Effective Capacitance

Effective capacitance is an essential concept when dealing with complex capacitive systems like those with multiple layers of dielectric materials. It helps simplify the problem of finding the total capacitance a certain configuration would exhibit.

- **Concept**: It refers to the single equivalent capacitance value that represents the entire arrangement of multiple capacitors (or dielectric materials) as one.- **Calculation**: To find the effective capacitance with dielectrics, we deduce a formula that combines the contributions of each dielectric. Deriving an expression for \( C_{eff} \), particularly in series arrangements, involves inversely summing up individual capacitances.- **Link to Dielectric Constant**: The effective dielectric constant \( K_{eff} \) is crucial for calculating effective capacitance for the whole ensemble. Simplifying the problem to a single capacitor with equivalent dielectric properties helps us understand the electric behavior of the arrangement.- **Result for the Exercise**: In our scenario, the effective constant \( K_{eff} \) was found using both the thickness and dielectric constants of the involved materials, leading to the expression \[ K = \frac{K_1 K_2 (d_1 + d_2)}{K_1 d_2 + K_2 d_1} \], aligning with choice (c). This provides a straightforward way to assess the system's overall electrical properties.