
What is the dimensional formula for electric field intensity?
$
{\text{A}}{\text{. }}\left[ {{\text{ML}}{{\text{T}}^{ - 3}}{{\text{A}}^{ - 1}}} \right] \\
{\text{B}}{\text{. }}\left[ {{\text{ML}}{{\text{T}}^{ - 1}}{{\text{A}}^{ - 3}}} \right] \\
{\text{C}}{\text{. }}\left[ {{\text{ML}}{{\text{T}}^3}{{\text{A}}^{ - 1}}} \right] \\
{\text{D}}{\text{. }}\left[ {{\text{ML}}{{\text{T}}^{ - 3}}{{\text{A}}^1}} \right] \\
$
Answer
232.8k+ views
Hint- Here, we will proceed by using the formulas for velocity i.e., Velocity = $\dfrac{{{\text{Length}}}}{{{\text{Time}}}}$, for acceleration i.e., Acceleration = $\dfrac{{{\text{Velocity}}}}{{{\text{Time}}}}$, for force i.e., Force = (Mass)$ \times $(Acceleration), for current i.e., Current = $\dfrac{{{\text{Magnitude of charge}}}}{{{\text{Time}}}}$ and for electric field intensity i.e., Electric field intensity = $\dfrac{{{\text{Force experienced}}}}{{{\text{Magnitude of charge}}}}$.
Complete step-by-step answer:
There are total seven fundamental quantities which exists and these are length whose dimension is denoted by [L], mass whose dimension is denoted by [M], time whose dimension is denoted by [T], temperature whose dimension is [K], electric current whose dimension is denoted by [A], luminous intensity whose dimension is denoted by [Cd] and amount of substance whose dimension is denoted by [mol].
All the other physical quantities (or their dimensions) can be represented in terms of these seven fundamental quantities (or their dimensions).
Since, Velocity = $\dfrac{{{\text{Length}}}}{{{\text{Time}}}}$
So, the dimension of velocity will be $\left[ {{\text{L}}{{\text{T}}^{ - 1}}} \right]$
Also, Acceleration = $\dfrac{{{\text{Velocity}}}}{{{\text{Time}}}}$
So, the dimension of acceleration will be $\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]$
Also, Force = (Mass)$ \times $(Acceleration)
So, the dimension of force experienced will be given as $\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]$
Also, Current = $\dfrac{{{\text{Magnitude of charge}}}}{{{\text{Time}}}}$
$ \Rightarrow $Magnitude of charge = (Current)$ \times $(Time)
So, the dimension of magnitude of charge will be $\left[ {{\text{AT}}} \right]$
Also, Electric field intensity = $\dfrac{{{\text{Force experienced}}}}{{{\text{Magnitude of charge}}}}$
So, the dimension of electric field intensity is given as $\dfrac{{\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]}}{{\left[ {{\text{AT}}} \right]}} = \left[ {{\text{ML}}{{\text{T}}^{ - 3}}{{\text{A}}^{-1}}} \right]$
Hence, option A is correct.
Note- In this particular problem, we have converted all the terms used in the formula for Electric field intensity = $\dfrac{{{\text{Force experienced}}}}{{{\text{Magnitude of charge}}}}$ in terms of the seven fundamental physical quantities because the dimensions of any physical quantity is given in terms of the dimensions of these fundamental quantities.
Complete step-by-step answer:
There are total seven fundamental quantities which exists and these are length whose dimension is denoted by [L], mass whose dimension is denoted by [M], time whose dimension is denoted by [T], temperature whose dimension is [K], electric current whose dimension is denoted by [A], luminous intensity whose dimension is denoted by [Cd] and amount of substance whose dimension is denoted by [mol].
All the other physical quantities (or their dimensions) can be represented in terms of these seven fundamental quantities (or their dimensions).
Since, Velocity = $\dfrac{{{\text{Length}}}}{{{\text{Time}}}}$
So, the dimension of velocity will be $\left[ {{\text{L}}{{\text{T}}^{ - 1}}} \right]$
Also, Acceleration = $\dfrac{{{\text{Velocity}}}}{{{\text{Time}}}}$
So, the dimension of acceleration will be $\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]$
Also, Force = (Mass)$ \times $(Acceleration)
So, the dimension of force experienced will be given as $\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]$
Also, Current = $\dfrac{{{\text{Magnitude of charge}}}}{{{\text{Time}}}}$
$ \Rightarrow $Magnitude of charge = (Current)$ \times $(Time)
So, the dimension of magnitude of charge will be $\left[ {{\text{AT}}} \right]$
Also, Electric field intensity = $\dfrac{{{\text{Force experienced}}}}{{{\text{Magnitude of charge}}}}$
So, the dimension of electric field intensity is given as $\dfrac{{\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]}}{{\left[ {{\text{AT}}} \right]}} = \left[ {{\text{ML}}{{\text{T}}^{ - 3}}{{\text{A}}^{-1}}} \right]$
Hence, option A is correct.
Note- In this particular problem, we have converted all the terms used in the formula for Electric field intensity = $\dfrac{{{\text{Force experienced}}}}{{{\text{Magnitude of charge}}}}$ in terms of the seven fundamental physical quantities because the dimensions of any physical quantity is given in terms of the dimensions of these fundamental quantities.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

