Problem 108 - Master Resource Book in JEE Main Physics

Question

A parallel plate capacitor has a capacitance of \(50 \mu \mathrm{F}\) in air and \(100 \mu \mathrm{F}\) when immersed in an oil. The dielectric constant \(K\) of the oil is (a) \(2.2\) (b) \(1.1\) (c) \(0.45\) (d) \(5.0\)

Step-by-Step Solution

Answer

Closest option is (a) 2.2, although calculation gives 2.

Step 1: Understanding Capacitance in Different Mediums

A parallel plate capacitor in air has a capacitance of \( C_0 = 50 \mu \mathrm{F} \). When immersed in a dielectric medium, the capacitance becomes \( C = 100 \mu \mathrm{F} \). The dielectric constant \( K \) is the ratio of the capacitance with the dielectric \( C \) to the capacitance in air \( C_0 \).

Step 2: Formula for Dielectric Constant

The formula to find the dielectric constant \( K \) is given by \[ K = \frac{C}{C_0} \]. This shows how much the capacitance increases when the dielectric is present.

Step 3: Substitute the Values

Substitute \( C = 100 \mu \mathrm{F} \) and \( C_0 = 50 \mu \mathrm{F} \) into the formula: \[ K = \frac{100 \mu \mathrm{F}}{50 \mu \mathrm{F}} \].

Step 4: Simplify the Expression

By simplifying the expression \( \frac{100}{50} \), we find that \( K = 2 \). This means the correct dielectric constant of the oil is \( 2 \).

Step 5: Correct Answer Selection

Comparing the calculated \( K = 2 \) with the given options, it does not exactly match any options. The problem may have a typo. The closest option \( a \) with \( K = 2.2 \) should be considered.

Key Concepts

Parallel Plate Capacitor

A parallel plate capacitor is a simple electrical component used to store energy in an electric field. It consists of two conductive plates separated by a distance, with an insulating material (dielectric) in between. The capacitor's ability to store charge per unit voltage is its capacitance. Here are some key features:

• Consists of two flat surfaces or plates.
• Plates are separated by an insulating layer (dielectric).
• Stores energy by holding opposite charges on each plate.

The capacitance, denoted as \( C \), is determined by the size of the plates, the separation distance between them, and the material (dielectric) filling the space. For a parallel plate capacitor, the basic formula for calculating capacitance is \[ C = \frac{\varepsilon A}{d} \]where:

• \( \varepsilon \): permittivity of the dielectric material.
• \( A \): area of one of the plates.
• \( d \): separation distance between the plates.

This formula highlights how increasing the plate area or reducing the separation distance increases the capacitor's overall capacitance.

Dielectric Constant

The dielectric constant, often symbolized as \( K \), is a measure of a material's ability to increase the capacitance of a capacitor compared to when a vacuum separates the capacitor's plates.

• It compares the capacitance of the capacitor with the dielectric to that without, in vacuum or air.
• A higher dielectric constant indicates a greater ability to store electrical energy within the field.

For any dielectric material, the relation is expressed through the formula: \[ K = \frac{C}{C_0} \]where:

• \( C \): Capacitance with the dielectric.
• \( C_0 \): Capacitance without the dielectric (in a vacuum or air).

Materials with high dielectric constants are used in capacitors to create efficient circuits that need to store large amounts of energy in small volumes. For instance, in the given exercise, the oil used as a dielectric has enhanced the capacitance from \( 50 \mu \mathrm{F} \) to \( 100 \mu \mathrm{F} \), resulting in a dielectric constant that significantly boosts the parallel plate capacitor's effectiveness.

Calculation of Capacitance

Calculating capacitance involves using the known dimensions of the capacitor, the properties of the dielectric material, and sometimes the dielectric constant. In our example, we calculated the dielectric constant using the formula:\[ K = \frac{C}{C_0} \]Let's break it down step by step:

• First, identify the capacitance without the dielectric, \( C_0 \). In our scenario, \( C_0 = 50 \mu \mathrm{F} \).
• Next, identify the capacitance with the dielectric, \( C = 100 \mu \mathrm{F} \).
• Substitute these values into the formula, giving us: \[ K = \frac{100 \mu \mathrm{F}}{50 \mu \mathrm{F}} = 2 \]

This demonstrates a simple, clear method to determine how a dielectric material, such as oil, impacts a capacitor's ability to store energy.