Match List - I with List - II. Choose the correct answer from the options given below:List - I List - II (A) Electric Field (I) [LTA] (B) Electric Flux (II) [L2] (C) Electric Dipole Moment (III) [ML3T−3A−1] (D) Area Vector Element (IV) [MLT−3A−1]
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Dimensional analysis is a powerful tool in physics that helps us understand the relationship between different physical quantities. Every physical quantity can be expressed in terms of fundamental dimensions like Mass (M), Length (L), Time (T), and Electric Current (A). The question asks us to match specific quantities from electromagnetism with their correct dimensional formulas. Let's analyze each quantity provided in List I.
We need to determine the dimensional formula for Electric Field, Electric Flux, Electric Dipole Moment, and Area Vector Element.
The electric field ($ \vec{E} $) is defined as the force experienced per unit charge. The formula is $ \vec{E} = \frac{\vec{F}}{q} $, where $ \vec{F} $ is the force and $ q $ is the charge.
The dimensional formula for Force is $ [\text{MLT}^{-2}] $. The dimensional formula for Charge is $ [\text{AT}] $ (since Current = Charge/Time, Charge = Current × Time).
So, the dimensional formula for Electric Field is:
\( [\text{E}] = \frac{[\text{F}]}{[\text{q}]} = \frac{[\text{MLT}^{-2}]}{[\text{AT}]} = [\text{MLT}^{-2}\text{A}^{-1}\text{T}^{-1}] = [\text{MLT}^{-3}\text{A}^{-1}] \)
This matches (IV) $ [\text{MLT}^{-3}\text{A}^{-1}] $ from List II.
Electric flux ($ \Phi_E $) represents the total number of electric field lines passing through a given area. It is defined by the integral $ \Phi_E = \int \vec{E} \cdot d\vec{A} $.
The dimensional formula for Electric Field is $ [\text{MLT}^{-3}\text{A}^{-1}] $, as derived above. The dimensional formula for Area ($ d\vec{A} $) is $ [\text{L}^2] $.
So, the dimensional formula for Electric Flux is:
\( [\Phi_E] = [\text{E}] \times [\text{Area}] = [\text{MLT}^{-3}\text{A}^{-1}] \times [\text{L}^2] = [\text{ML}^{1+2}\text{T}^{-3}\text{A}^{-1}] = [\text{ML}^3\text{T}^{-3}\text{A}^{-1}] \)
This matches (III) $ [\text{ML}^3\text{T}^{-3}\text{A}^{-1}] $ from List II.
An electric dipole moment ($ \vec{p} $) is defined as the product of the magnitude of one of the charges ($ q $) and the distance ($ d $) between the two charges forming the dipole. The formula is $ \vec{p} = q\vec{d} $.
The dimensional formula for Charge ($ q $) is $ [\text{AT}] $. The dimensional formula for distance ($ d $) is $ [\text{L}] $.
So, the dimensional formula for Electric Dipole Moment is:
\( [\text{p}] = [\text{q}] \times [\text{d}] = [\text{AT}] \times [\text{L}] = [\text{LTA}] \)
This matches (I) $ [\text{LTA}] $ from List II.
An area vector element ($ d\vec{A} $) represents an infinitesimal area. Area is a measure of the extent of a two-dimensional surface.
The dimensional formula for any Area is simply the square of the dimension of Length.
\( [\text{Area}] = [\text{L}]^2 = [\text{L}^2] \)
This matches (II) $ [\text{L}^2] $ from List II.
Based on the dimensional analysis, the correct matching is:
Comparing this matching with the given options:
The correct option is the one that matches our derived pairings.
| List - I (Quantity) | Dimensional Formula | List - II Match |
|---|---|---|
| (A) Electric Field ($ \vec{E} $) | $ [\text{MLT}^{-3}\text{A}^{-1}] $ | (IV) |
| (B) Electric Flux ($ \Phi_E $) | $ [\text{ML}^3\text{T}^{-3}\text{A}^{-1}] $ | (III) |
| (C) Electric Dipole Moment ($ \vec{p} $) | $ [\text{LTA}] $ | (I) |
| (D) Area Vector Element ($ d\vec{A} $) | $ [\text{L}^2] $ | (II) |
The correct correspondence is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).
| Quantity | Common Symbol | Derivation Basis | Dimensional Formula (M, L, T, A) |
|---|---|---|---|
| Charge | \( q \) | Current \( \times \) Time | \( [\text{AT}] \) |
| Electric Field | \( \vec{E} \) | Force / Charge | \( [\text{MLT}^{-3}\text{A}^{-1}] \) |
| Electric Flux | \( \Phi_E \) | Electric Field \( \times \) Area | \( [\text{ML}^3\text{T}^{-3}\text{A}^{-1}] \) |
| Electric Dipole Moment | \( \vec{p} \) | Charge \( \times \) Distance | \( [\text{LTA}] \) |
| Area | \( A \) | Length \( \times \) Length | \( [\text{L}^2] \) |
| Force | \( \vec{F} \) | Mass \( \times \) Acceleration | \( [\text{MLT}^{-2}] \) |
Dimensional analysis is crucial in physics for several reasons:
By expressing quantities in terms of fundamental dimensions, we can simplify problems and gain deeper insight into the physical nature of equations and concepts.
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A thin metallic spherical shell contains a charge +10 μC on it. A point charge +2 μC is placed at the centre of the shell and another charge +5 μC is placed outside it as shown. The force on the charge +2 μC at the centre is:

In the figure, an α-particle moves a distance l in a uniform electric field E as shown. Does the Electric Field do a positive or a negative work on the α-particle? Does the electric potential energy of the α-particle increase or decrease?

The force between two electric charges is expressed by the equation:
F = (k q1 q2) / r2
Which of the following is a correct statement?