Problem 79 - Master Resource Book in JEE Main Physics

Question

A parallel plate capacitor has the space between its plates filled by two slabs of thickness \(\frac{d}{2}\) each and dielectric constants \(K_{1}\) and \(K_{2} . d\) is the plate separation of the capacitor. The capacity of the capacitor is (a) \(\frac{2 \varepsilon_{0} d}{A}\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right)\) (b) \(\frac{2 \varepsilon_{0} A}{d}\left(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right)\) (c) \(\frac{2 \varepsilon_{0} d}{A}\left(K_{1}+K_{2}\right)\) (d) \(\frac{2 \varepsilon_{\mathrm{o}} A}{d}\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right)\)

Step-by-Step Solution

Answer

The correct option is (b): \(\frac{2 \varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1+K_2}\right)\).

Step 1: Understanding Dielectric Capacitors

To find the capacitance of a parallel plate capacitor with dielectric slabs, we need to consider how dielectrics alter the effective capacitance. Dielectrics increase the capacitance by a factor equal to their dielectric constant.

Step 2: Analyzing Slabs Arrangement

The capacitor has two slabs, each of thickness \(\frac{d}{2}\), with dielectric constants \(K_1\) and \(K_2\). Since the slabs are aligned in the direction of the electric field, calculate the resultant capacitance using their series combination.

Step 3: Capacitance of a Single Slab

For each slab, calculate capacitance using the formula for a dielectric filled capacitor: \(C = \frac{\varepsilon_0 A}{d'} K\), where \(d' = \frac{d}{2}\). Thus, \(C_1 = \frac{2 \varepsilon_0 A}{d} K_1\) and \(C_2 = \frac{2 \varepsilon_0 A}{d} K_2\).

Step 4: Effective Capacitance in Series

The total capacitance, \(C\), for capacitors in series is given by: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]. Substituting \(C_1\) and \(C_2\) gives \( \frac{1}{C} = \frac{d}{2 \varepsilon_0 A K_1} + \frac{d}{2 \varepsilon_0 A K_2} \).

Step 5: Simplifying the Expression

Combine the expressions for \(\frac{1}{C_1}\) and \(\frac{1}{C_2}\) and simplify: \( \frac{1}{C} = \frac{d (K_1 + K_2)}{2 \varepsilon_0 A K_1 K_2} \). Thus, \( C = \frac{2 \varepsilon_0 A K_1 K_2}{d (K_1 + K_2)} \).

Step 6: Identify the Correct Option

Comparing with the options, \( C = \frac{2 \varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1+K_2} \right) \), which corresponds to option (b).

Key Concepts

Dielectric Constant

When studying capacitors, one essential property to understand is the dielectric constant, often represented as \( K \). This constant is pivotal because it measures a material's ability to store electrical energy in an electric field. The dielectric constant indicates how much more capacity a dielectric material contributes to a capacitor compared to a vacuum or air. For a material with a dielectric constant \( K \), the capacitance increases by a multiplication factor of \( K \).

Essentially, a dielectric material inserted between the plates of a capacitor boosts its capacitance by reducing the electric field within the capacitor. The dielectric constant is a dimensionless number that tells us how effectively the dielectric material can increase the efficiency of a capacitor, allowing it to hold more charge for a given voltage.

Series Capacitance

In capacitors, series connections result in a change of the overall capacitance unlike parallel arrangements. When capacitors are in series, their effective capacitance is less than any single capacitor in the connection. For a series of capacitors filled with dielectric materials with varying dielectric constants, you compute the effective capacitance using the following relation:

• The formula for series capacitance is \( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \)
• In the context of our exercise, the slabs are treated as capacitors with individual capacitances \( C_1 \) and \( C_2 \).

Each slab alters the overall capacitance in the same fashion as a capacitor in a simple series circuit. It's crucial to combine and simplify these values accurately, as outlined, to find the resultant effect or the effective capacitance of the entire configuration.

Effective Capacitance

The effective capacitance of a system with dielectric materials can be understood as the agent representing the total storage capability of the capacitor system. In scenarios with multiple dielectric slabs, such as the capacitor with slabs described in the exercise, the term "effective capacitance" becomes vital.

To find the effective capacitance, consider all the capacitive effects of the slabs in series. You first find individual capacitances based on the dielectric measurements and individual configurations, then apply the formula for series capacitance to bridge these values together:

• The expression \( \frac{1}{C} = \frac{d (K_1 + K_2)}{2 \varepsilon_0 A K_1 K_2} \) shows how to compute it.
• After simplifying, it results in the effective capacitance formula: \( C = \frac{2 \varepsilon_0 A K_1 K_2}{d (K_1 + K_2)} \).

This process leads to the understanding that, although each component might have differing dielectric effects, collectively, they contribute to an effective capacitance that helps us gauge how much potential charge the entire system can handle. This calculated effective capacitance corresponds to option (b) in the exercise, which is the correct answer.