Problem 57 - Master Resource Book in JEE Main Physics

Question

The dimensions of permittivity \(\varepsilon_{0}\) are \(\quad[B V P 2006]\) (a) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~A}^{2} \mathrm{~T}^{4}\right]\) (b) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~A}^{-2} \mathrm{~T}^{-4}\right]\) (c) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~A}^{2} \mathrm{~T}^{2}\right]\) (d) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~A}^{2} \mathrm{~T}^{-4}\right]\)

Step-by-Step Solution

Answer

The dimensions of \( \varepsilon_0 \) are \([M^{-1} L^{-3} A^2 T^4]\), which matches option (a).

Step 1: Understand the Problem

We are tasked with finding the dimensions of permittivity \( \varepsilon_0 \). This involves expressing \( \varepsilon_0 \) in terms of base physical dimensions such as Mass \(M\), Length \(L\), Time \(T\), and Electric Current \(A\).

Step 2: Recall the Relevant Equations

The permittivity \( \varepsilon_0 \) appears in Coulomb's Law as \( F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \), where \( F \) is force, \( q_1 \) and \( q_2 \) are charges, and \( r \) is the distance between the charges. Also, permittivity is related to the capacitance \( C \) of a capacitor by \( C = \varepsilon_0 \frac{A}{d} \), where \( A \) is the area and \( d \) is the separation between the plates.

Step 3: Use Dimensions from Coulomb's Law

In Coulomb's Law, \( F \) has dimensions \([MLT^{-2}]\), charge \( q \) has dimensions \([AT]\), and distance \( r \) has dimensions \([L]\). Solving \( \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} = F \) for \( \varepsilon_0 \) gives us \( [\varepsilon_0] = \frac{[q]^2}{[F][r]^2} = \frac{[A^2T^2]}{[MLT^{-2}][L^2]} = [M^{-1}L^{-3}A^2T^4] \).

Step 4: Match the Dimensions

From Step 3, we found that the dimensions of \( \varepsilon_0 \) are \([M^{-1} L^{-3} A^2 T^4]\). Comparing with the given options, we see that this matches option (a) \([M^{-1} L^{-3} A^2 T^4]\).

Key Concepts

Permittivity

Permittivity, denoted as \( \varepsilon \), is a fundamental property in electromagnetism that affects how electric fields interact with materials. In a vacuum, this value is referred to as the electric constant or \( \varepsilon_{0} \), with a value approximately equal to \( 8.85 \times 10^{-12} \, \text{F/m} \) (farads per meter). This parameter is crucial as it influences the force between two charged objects.

Permittivity can be understood as a measure of the resistance that is encountered when forming an electric field within a medium. Different materials have different permittivity values, leading to diverse electrical behaviors.

• A higher permittivity indicates that a material can store more electrical energy within an electric field.
• A lower permittivity suggests that the material stores less energy.

The dimensional formula for permittivity can be derived based on its involvement in electromagnetic principles such as Coulomb's law.

Understanding the dimensions of permittivity \( [M^{-1}L^{-3}A^2T^4] \) allows you to connect theoretical physics to measurable quantities.

Coulomb's Law

Coulomb's Law is a pivotal principle in electrostatics, describing the force between two point charges. It is stated as: \[ F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \]where:

• \( F \) is the electrostatic force between charges.
• \( q_1 \) and \( q_2 \) are magnitudes of these charges.
• \( r \) is the separation distance between the charges.

The law highlights that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The constant \( \varepsilon_0 \) acts here as a scaling factor, tuning this force according to the medium.

To determine the influence of electric permittivity, we look at its dimensional involvement, reinforcing how varying materials can change interaction strengths. And through dimensional analysis, we see how fundamental quantities like mass, length, time, and electric current relate to permittivity within Coulomb’s Law.

Capacitance

Capacitance is the capacity of a system to store electrical energy in the form of an electric charge. Represented by the letter \( C \), it is fundamentally linked to the ability of a capacitor to resist changes in voltage.

The relationship between capacitance and permittivity is given by:\[ C = \varepsilon_0 \frac{A}{d} \]Here:

• \( C \) stands for capacitance measured in farads (F).
• \( A \) is the area of overlap of the two plates in a capacitor.
• \( d \) is the separation between the plates.

Capacitance denotes the amount of charge required per unit voltage to increase the potential across the plates by one volt. The permittivity \( \varepsilon_0 \) acts as a proportionality factor affecting the total energy storage capacity based on the medium between the plates.

Capacitance also integrates the dimensions of permittivity, influencing the design of electronic components like capacitors that are essential in managing energy storage and distribution in electrical circuits. By understanding these relationships, we gain insight into the fundamental operations of electronics.