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Question: Electric displacement is given by \(D = \varepsilon E\) Here, \(\varepsilon = \) electric permitti...

Electric displacement is given by D=εED = \varepsilon E
Here, ε=\varepsilon = electric permittivity
E=E = electric field strength
The dimensions of electric displacement is:
(A) [ML−2TA]\left[ {M{L^{ - 2}}TA} \right]
(B) [L−2T−1A]\left[ {{L^{ - 2}}{T^{ - 1}}A} \right]
(C) [L−2TA]\left[ {{L^{ - 2}}TA} \right]
(D) None of these

Explanation

Solution

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=Qq4πε0r2F = \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 QQ we get the electric field as
E=q4πε0r2E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}
Multiplying by the electrical permittivity ε0{\varepsilon _0} both the sides, we get
ε0E=q4πr2{\varepsilon _0}E = \dfrac{q}{{4\pi {r^2}}} ……………..(1)
According to the question, the electric displacement is given by
D=εED = \varepsilon E …………………….(2)
Substituting (1) in (2) we get the electric displacement as
D=q4πr2D = \dfrac{q}{{4\pi {r^2}}}
Now, taking the dimensions of both the sides, we get
[D]=[q4πr2]\left[ D \right] = \left[ {\dfrac{q}{{4\pi {r^2}}}} \right]
Since 4π4\pi is a constant, it is dimensionless. Therefore we get
[D]=[q][r]2\left[ D \right] = \dfrac{{\left[ q \right]}}{{{{\left[ r \right]}^2}}} ………………….(3)
Now, we know that the dimensions of the charge are given by
[q]=[AT]\left[ q \right] = \left[ {AT} \right] ………………..(4)
Also, the dimensions of the distance are given by
[r]=[L]\left[ r \right] = \left[ L \right] ……………………..(5)
Substituting (4) and (5) in (3) we get
[D]=[AT][L]2\left[ D \right] = \dfrac{{\left[ {AT} \right]}}{{{{\left[ L \right]}^2}}}
On simplifying, we finally get
[D]=[L−2TA]\left[ D \right] = \left[ {{L^{ - 2}}TA} \right]
The dimensions of the electric displacement are [L−2TA]\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.