Match List I with List II.

List I
A. Capacitance, C
B. Permittivity of free space, e0
C. Permeabili

Q: Why do we use Charge ($A \cdot T$) instead of just Current ($A$)?

In the SI system, Current ($A$) is the fundamental quantity, not Charge. However, since charge $q = I \cdot t$, its dimensional formula is $[AT]$. It's often easier to substitute $[AT]$ directly whenever you see a charge $q$ in an equation.

Q: Can I find the dimensions of $\mu_0$ using the magnetic force formula?

Absolutely! You can use the force per unit length between two parallel wires: $F/L = (\mu_0 I_1 I_2) / (2\pi d)$. Rearranging for $\mu_0$ gives the exact same dimensional formula, but using the speed of light equation is often faster if you already know $\varepsilon_0$.

Q: Why is work used to find the dimensions of capacitance?

Capacitance is $C = q/V$. Potential $V$ is defined as work done per unit charge ($W/q$). Substituting this makes $C = q^2/W$, which is much easier to break down into fundamental dimensions since we all know work is $[ML^2T^{-2}]$.

Q: What if I forget the dimensional formula for Force?

Force is mass times acceleration ($F = ma$). Mass is $[M]$, and acceleration is velocity over time, or distance over time squared ($[LT^{-2}]$). So, Force is always $[MLT^{-2}]$. This is the most crucial dimensional formula to memorize!

Q: Is there a way to verify if my dimensional formula is correct?

Yes! You can cross-check by using a completely different formula that contains the same quantity. For example, for electric field, you could use $E = V/d$ (potential gradient) instead of $E = F/q$. Both will yield $[M^1L^1T^{-3}A^{-1}]$.

Question

Match List I with List II.

\begin{array}{ll} 	ext{List I} & 	ext{List II} \ 	ext{A. Capacitance, } C & 	ext{I. } M^1 L^1 T^{-3} A^{-1} \ 	ext{B. Permittivity of free space, } \varepsilon_0 & 	ext{II. } M^{-1} L^{-3} T^4 A^2 \ 	ext{C. Permeability of free space, } \mu_0 & 	ext{III. } M^{-1} L^{-2} T^4 A^2 \ 	ext{D. Electric field, } E & 	ext{IV. } M^1 L^1 T^{-2} A^{-2} \end{array}

Choose the correct answer from the options given below.

Options:

(A) A $ightarrow$ III, B $ightarrow$ II, C $ightarrow$ IV, D $ightarrow$ I

(B) A $ightarrow$ III, B $ightarrow$ IV, C $ightarrow$ II, D $ightarrow$ I

(C) A $ightarrow$ IV, B $ightarrow$ II, C $ightarrow$ III, D $ightarrow$ I

(D) A $ightarrow$ IV, B $ightarrow$ III, C $ightarrow$ II, D $ightarrow$ I

Sigma Percentile · Accepted Answer

### The Beauty of Dimensional Analysis

Imagine you are an explorer in the vast universe of physics, and your most trusted compass is Dimensional Analysis. It is the ultimate truth-teller, the universal translator that strips away the complexity of any physical quantity and reveals its fundamental DNA: Mass ($M$), Length ($L$), Time ($T$), and Current ($A$).

In this epic problem, we are tasked with decoding four of the most profound quantities in electromagnetism: Capacitance, Permittivity, Permeability, and the Electric Field. Let's break them down one by one.

### Decoding Capacitance ($C$)

Capacitance is the ability of a system to store electric charge. Mathematically, it is defined as the charge stored per unit potential difference:

$$C = \frac{q}{V}$$

But what is potential ($V$)? Potential is simply the work done per unit charge ($V = W/q$). Substituting this into our capacitance equation gives us a much friendlier form:

$$C = \frac{q^2}{W}$$

Now, we just plug in the fundamental dimensions. Charge $q$ is current times time, so $[q] = [AT]$. Work $W$ is force times distance, so $[W] = [ML^2T^{-2}]$.

$$[C] = \frac{[AT]^2}{[ML^2T^{-2}]} = [M^{-1}L^{-2}T^4A^2]$$

This perfectly matches option **III** in our list.

### Unveiling Permittivity ($\varepsilon_0$)

Permittivity of free space tells us how much resistance a vacuum offers to the formation of an electric field. The easiest way to find its dimensions is to summon Coulomb's Law:

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$$

By isolating $\varepsilon_0$, we get:

$$\varepsilon_0 = \frac{q^2}{4\pi F r^2}$$

We know the dimensions of force ($[MLT^{-2}]$), charge ($[AT]$), and distance ($[L]$). Let's substitute them in:

$$[\varepsilon_0] = \frac{[AT]^2}{[MLT^{-2}][L^2]} = [M^{-1}L^{-3}T^4A^2]$$

This matches option **II**.

### The Speed of Light Trick for Permeability ($\mu_0$)

Permeability of free space is the magnetic counterpart to permittivity. While you could use the Biot-Savart law or Ampere's law to find its dimensions, there is a much more elegant trick. James Clerk Maxwell showed us that the speed of light in a vacuum is intimately connected to these two constants:

$$c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$$

Squaring both sides and rearranging for $\mu_0$ gives:

$$\mu_0 = \frac{1}{\varepsilon_0 c^2}$$

We already did the hard work of finding $[\varepsilon_0]$. The speed of light $c$ is just a velocity, so $[c] = [LT^{-1}]$. Let's plug these in:

$$[\mu_0] = \frac{1}{[M^{-1}L^{-3}T^4A^2][LT^{-1}]^2}$$

Simplifying the denominator and bringing the terms to the numerator yields:

$$[\mu_0] = [M^1L^1T^{-2}A^{-2}]$$

This matches option **IV**.

### The Simplicity of the Electric Field ($E$)

Finally, we arrive at the Electric Field. An electric field is simply the electrostatic force experienced per unit charge:

$$E = \frac{F}{q}$$

This is the most straightforward calculation of the bunch. We divide the dimensions of force by the dimensions of charge:

$$[E] = \frac{[MLT^{-2}]}{[AT]} = [M^1L^1T^{-3}A^{-1}]$$

This matches option **I**.

### Bringing It All Together

By systematically breaking down each complex quantity into fundamental mechanical dimensions, we have successfully decoded the entire matrix:

- **A. Capacitance** $\rightarrow$ **III**
- **B. Permittivity** $\rightarrow$ **II**
- **C. Permeability** $\rightarrow$ **IV**
- **D. Electric Field** $\rightarrow$ **I**

This perfectly aligns with the first option provided in the question. Dimensional analysis is not just a tool for solving problems; it is a profound way to see the underlying unity of the physical world.

Select Answer:

Visualized Solution

$Dimensional Analysis of EM Quantities$

$Capacitance (C)$

$Dimensions of C$

$Permittivity (ε_{0})$

$Dimensions of ε_{0}$

$Permeability (μ_{0})$

$Dimensions of μ_{0}$

$Electric Field (E)$

$Dimensions of E$

$Final Matching$

$The Way Forward$

The Sigma Insight: Dimensional Analysis

The Beauty of Dimensional Analysis

Decoding Capacitance ( $C$ )

Unveiling Permittivity ( $ε_{0}$ )

The Speed of Light Trick for Permeability ( $μ_{0}$ )

The Simplicity of the Electric Field ( $E$ )

Bringing It All Together

Similar Questions

Comprehension Passage

The relation between $[ε_{0}]$ and $[μ_{0}]$ is

The dimensions of $\frac{1}{2} ε_{0} E^{2}$ ( $ε_{0}$ : permittivity of free space; $E$ : electric field) is

Let $[ε_{0}]$ denote the dimensional formula of the permittivity of the vacuum and $[μ_{0}]$ that of the permeability of the vacuum. If $M = mass$ , $L = length$ , $T = time$ and $I = electric current$ .

Match List-I with List-II Choose the correct answer from the options given below.

List-I

List-II

Let $[ε_{0}]$ denotes the dimensional formula of the permittivity of vacuum. If $M = mass$ , $L = length$ , $T = time$ and $A = electric current$ , then

Comprehension Passage

The relation between $[E]$ and $[B]$ is

Match the physical quantities given in Column I with dimensions expressed in terms of mass ( $M$ ), length ( $L$ ), time ( $T$ ), and charge ( $Q$ ) given in Column II and write the correct answer against the matched quantity in a tabular form in your answer book.

List-I

List-II

A quantity $X$ is given by $ε_{0} L \frac{Δ V}{Δ t}$ , where $ε_{0}$ is the permittivity of free space, $L$ is a length, $Δ V$ is a potential difference and $Δ t$ is a time interval. The dimensional formula for $X$ is the same as that of

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensional Analysis of EM Quantities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Capacitance (C)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensions of C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Permittivity (ε0​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensions of ε0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Permeability (μ0​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensions of μ0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field (E)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dimensions of E

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Matching

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward

The Sigma Insight: Dimensional Analysis

The Beauty of Dimensional Analysis

Decoding Capacitance ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C)

Unveiling Permittivity ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ε0​)

The Speed of Light Trick for Permeability ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ0​)

The Simplicity of the Electric Field ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E)

Bringing It All Together

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The relation between [ε0​] and [μ0​] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The dimensions of 21​ε0​E2 (ε0​ : permittivity of free space; E : electric field) is

Match List-I with List-II Choose the correct answer from the options given below.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let [ε0​] denotes the dimensional formula of the permittivity of vacuum. If M=mass, L=length, T=time and A=electric current, then

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The relation between [E] and [B] is

List-I

List-II

$Dimensional Analysis of EM Quantities$

$Capacitance (C)$

$Dimensions of C$

$Permittivity (ε_{0})$

$Dimensions of ε_{0}$

$Permeability (μ_{0})$

$Dimensions of μ_{0}$

$Electric Field (E)$

$Dimensions of E$

$Final Matching$

$The Way Forward$

Decoding Capacitance ( $C$ )

Unveiling Permittivity ( $ε_{0}$ )

The Speed of Light Trick for Permeability ( $μ_{0}$ )

The Simplicity of the Electric Field ( $E$ )

The relation between $[ε_{0}]$ and $[μ_{0}]$ is

The dimensions of $\frac{1}{2} ε_{0} E^{2}$ ( $ε_{0}$ : permittivity of free space; $E$ : electric field) is

Let $[ε_{0}]$ denotes the dimensional formula of the permittivity of vacuum. If $M = mass$ , $L = length$ , $T = time$ and $A = electric current$ , then

The relation between $[E]$ and $[B]$ is