
Electric displacement is given by $D = \varepsilon E$
Here, $\varepsilon = $ electric permittivity
$E = $ electric field strength
The dimensions of electric displacement is:
(A) $\left[ {M{L^{ - 2}}TA} \right]$
(B) $\left[ {{L^{ - 2}}{T^{ - 1}}A} \right]$
(C) $\left[ {{L^{ - 2}}TA} \right]$
(D) None of these
Answer
174.3k+ views
Hint: To solve this question, we have to use the Coulomb’s formula of the electrostatic force between two charged particles. From there we can get the expression of the electric displacement in terms of the charge and the distance, from which its dimensions can be deduced.
Complete step-by-step answer:
We know from the Coulomb’s law that the electrostatic force between two charges is given by
$F = \dfrac{{Qq}}{{4\pi {\varepsilon _0}{r^2}}}$
Also, we know that the electric field strength is equal to the electrostatic force per unit charge. So dividing both sides of the above equation with the charge $Q$ we get the electric field as
$E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}$
Multiplying by the electrical permittivity ${\varepsilon _0}$ both the sides, we get
${\varepsilon _0}E = \dfrac{q}{{4\pi {r^2}}}$ ……………..(1)
According to the question, the electric displacement is given by
$D = \varepsilon E$ …………………….(2)
Substituting (1) in (2) we get the electric displacement as
$D = \dfrac{q}{{4\pi {r^2}}}$
Now, taking the dimensions of both the sides, we get
$\left[ D \right] = \left[ {\dfrac{q}{{4\pi {r^2}}}} \right]$
Since \[4\pi \] is a constant, it is dimensionless. Therefore we get
\[\left[ D \right] = \dfrac{{\left[ q \right]}}{{{{\left[ r \right]}^2}}}\] ………………….(3)
Now, we know that the dimensions of the charge are given by
$\left[ q \right] = \left[ {AT} \right]$ ………………..(4)
Also, the dimensions of the distance are given by
$\left[ r \right] = \left[ L \right]$ ……………………..(5)
Substituting (4) and (5) in (3) we get
\[\left[ D \right] = \dfrac{{\left[ {AT} \right]}}{{{{\left[ L \right]}^2}}}\]
On simplifying, we finally get
$\left[ D \right] = \left[ {{L^{ - 2}}TA} \right]$
The dimensions of the electric displacement are $\left[ {{L^{ - 2}}TA} \right]$.
Hence, the correct answer is option C.
Note: We could also separately find out the dimensions of the electrical permittivity and the electric field to get the dimensions of the electric displacement. But that would involve time taking calculations, so we directly found out the expression for the electric displacement in terms of the simple parameters.
Complete step-by-step answer:
We know from the Coulomb’s law that the electrostatic force between two charges is given by
$F = \dfrac{{Qq}}{{4\pi {\varepsilon _0}{r^2}}}$
Also, we know that the electric field strength is equal to the electrostatic force per unit charge. So dividing both sides of the above equation with the charge $Q$ we get the electric field as
$E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}$
Multiplying by the electrical permittivity ${\varepsilon _0}$ both the sides, we get
${\varepsilon _0}E = \dfrac{q}{{4\pi {r^2}}}$ ……………..(1)
According to the question, the electric displacement is given by
$D = \varepsilon E$ …………………….(2)
Substituting (1) in (2) we get the electric displacement as
$D = \dfrac{q}{{4\pi {r^2}}}$
Now, taking the dimensions of both the sides, we get
$\left[ D \right] = \left[ {\dfrac{q}{{4\pi {r^2}}}} \right]$
Since \[4\pi \] is a constant, it is dimensionless. Therefore we get
\[\left[ D \right] = \dfrac{{\left[ q \right]}}{{{{\left[ r \right]}^2}}}\] ………………….(3)
Now, we know that the dimensions of the charge are given by
$\left[ q \right] = \left[ {AT} \right]$ ………………..(4)
Also, the dimensions of the distance are given by
$\left[ r \right] = \left[ L \right]$ ……………………..(5)
Substituting (4) and (5) in (3) we get
\[\left[ D \right] = \dfrac{{\left[ {AT} \right]}}{{{{\left[ L \right]}^2}}}\]
On simplifying, we finally get
$\left[ D \right] = \left[ {{L^{ - 2}}TA} \right]$
The dimensions of the electric displacement are $\left[ {{L^{ - 2}}TA} \right]$.
Hence, the correct answer is option C.
Note: We could also separately find out the dimensions of the electrical permittivity and the electric field to get the dimensions of the electric displacement. But that would involve time taking calculations, so we directly found out the expression for the electric displacement in terms of the simple parameters.
Recently Updated Pages
JEE Main 2025-26 Atoms and Nuclei Mock Test: Free Practice Online

JEE Main 2025-26: Dual Nature of Matter and Radiation Mock Test

JEE Main 2025-26 Electronic Devices Mock Test – Free Practice

JEE Main Mock Test 2025-26: Experimental Skills Chapter Online Practice

JEE Main 2025-26 Current Electricity Mock Test: Free Practice Online

JEE Main 2025-26 Rotational Motion Mock Test – Free Practice Online

Trending doubts
Uniform Acceleration

Instantaneous Velocity - Formula based Examples for JEE

Electron Gain Enthalpy and Electron Affinity for JEE

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Current Loop as Magnetic Dipole and Its Derivation for JEE

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

JEE Advanced 2025 Notes

The ratio of the lengths densities masses and resistivities class 12 physics JEE_Main
