The dimension of permittivity $(\varepsilon _0)$ are ______. Take Q as the dimension of charge.
A) $[M^{-1} L^{-2} T^{-2} Q^{-2}]$
B) $[M^{-1} L^{-3} T^{2} Q^{2}]$
C) $[M^{-1} L^{2} T^{-3} Q^{-1}]$
D) $[M^{-1} L^{3} T^{-2} Q^{-2}]$
Answer
300.6k+ views
Hint: The dimensions of a physical quantity are the power raised by the basic units to obtain a unit of those quantities. For derived quantity represented by \[{Q = }{{{M}}^{{a}}}{{{L}}^{{b}}}{{{T}}^{{c}}}\], then \[{{{M}}^{{a}}}{{{L}}^{{b}}}{{{T}}^{{c}}}\] is called dimensional formula.
Complete solution:
${F = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
and dimension of $F = [M L T^{-2}]$
$\Rightarrow {ML}{{{T}}^{-2}}{ = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
$\Rightarrow K = [M L^3 T^{-2} Q^{-2}]$
Substituting value of K
$\Rightarrow \dfrac{1}{4 \pi \varepsilon _0}$
Hence, answer is option (B), $[M^{-1} L^{-3} T^2 Q^2]$.
Additional Information:
Permeability and dielectric constant are two terms that are central to capacitor technology. It is often talked about that capacitors will be used with different dielectrics. All common names for electrolytic capacitors, ceramic capacitors, paper, tantalum capacitors, and capacitors refer to the dielectric material used.
The dielectric components provide insulation between the capacitor plates and in addition it determines many of the properties of the capacitor. This capacitance is achieved in constant quantities, whether the temperature constant is polarized or not. These and many other properties are controlled by the dielectric material used - many properties are managed only by the dielectric constant.
Note: Vacuum refers to the minimum possible value of permittivity allowed. This is commonly known as free space or permitted electronic constant. Denoted by \[ \varepsilon _0\] and has the value \[8.85 \times {10}^{-12} Farad/meter\]. Disputes against the formation of electric field lines are also evident in the indictment. The permittivity of a dielectric is represented by the ratio of its absolute permittivity to the electronic constant and is usually given relative to the erase.
Complete solution:
${F = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
and dimension of $F = [M L T^{-2}]$
$\Rightarrow {ML}{{{T}}^{-2}}{ = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
$\Rightarrow K = [M L^3 T^{-2} Q^{-2}]$
Substituting value of K
$\Rightarrow \dfrac{1}{4 \pi \varepsilon _0}$
Hence, answer is option (B), $[M^{-1} L^{-3} T^2 Q^2]$.
Additional Information:
Permeability and dielectric constant are two terms that are central to capacitor technology. It is often talked about that capacitors will be used with different dielectrics. All common names for electrolytic capacitors, ceramic capacitors, paper, tantalum capacitors, and capacitors refer to the dielectric material used.
The dielectric components provide insulation between the capacitor plates and in addition it determines many of the properties of the capacitor. This capacitance is achieved in constant quantities, whether the temperature constant is polarized or not. These and many other properties are controlled by the dielectric material used - many properties are managed only by the dielectric constant.
Note: Vacuum refers to the minimum possible value of permittivity allowed. This is commonly known as free space or permitted electronic constant. Denoted by \[ \varepsilon _0\] and has the value \[8.85 \times {10}^{-12} Farad/meter\]. Disputes against the formation of electric field lines are also evident in the indictment. The permittivity of a dielectric is represented by the ratio of its absolute permittivity to the electronic constant and is usually given relative to the erase.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

Understanding Uniform Acceleration in Physics

Hybridisation in Chemistry – Concept, Types & Applications

Effective Nuclear Charge for JEE

Understanding Elastic Collisions in Two Dimensions

