Question

The dimension of $\left( \dfrac{1}{2} \right){{\varepsilon }_{0}}{{E}^{2}}$ is (${{\varepsilon }_{0}}$ : permittivity of free space, E: Electric field)
A. $ML{{T}^{-1}}$
B. $M{{L}^{2}}{{T}^{-2}}$
C. $M{{L}^{-1}}{{T}^{-2}}$
D. $M{{L}^{2}}{{T}^{-1}}$

Answer 1

To answer this question, we are to express the above quantity in combinations of seven fundamental quantities that we know. We could separate the dimensions of the electric field from its very definition and that of permittivity of free space from Coulomb's law.

Formula used:
Coulomb’s law,
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Definition of electric field,
$E=\dfrac{F}{q}$

Complete step-by-step answer:
We know that any physical quantity can be expressed in terms of a particular combination of the seven known fundamental quantities like length [L], mass [M], time [T], electric current[A], thermodynamic temperature[K], luminous intensity[cd] and amount of substance[mol].
To define the dimension of a physical quantity, we could say they are the powers to which the base quantities are raised to represent that quantity.
Here we are asked to find the dimension of $\left( \dfrac{1}{2} \right){{\varepsilon }_{0}}{{E}^{2}}$
Let us separately find the dimensions of the electric field and permittivity of free space.
We can derive the dimension of Permittivity of free space from the Coulomb’s law excluding the dimensionless constant (these are constants with no associated units),
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ ………………. (1)
We have F=ma from Newton’s second law,
$force=mass\times \dfrac{length}{tim{{e}^{2}}}$
So the dimension of force is,
$\left[ F \right]=\left[ ML{{T}^{-2}} \right]$ ……………………. (2)
Charge q is the product of current and time so its dimension becomes,
$\left[ q \right]=\left[ AT \right]$ …………………………… (3)
Substituting (2) and (3) for the dimensions of the respective quantities in equation (1) we get,
$\left[ {{\varepsilon }_{0}} \right]=\dfrac{{{\left[ q \right]}^{2}}}{\left[ F \right]{{\left[ r \right]}^{2}}}$
$\left[ {{\varepsilon }_{0}} \right]=\dfrac{{{\left[ AT \right]}^{2}}}{\left[ ML{{T}^{-2}} \right]{{\left[ L \right]}^{2}}}$
$\left[ {{\varepsilon }_{0}} \right]=\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]$ ………………………… (4)
From the definition of electric field we know that it is the force per unit charge, that is,
$E=\dfrac{F}{q}$
Dimension of the electric field can thus be derived from this relation by substituting (2) and (3) for the dimensions of force and charge respectively. Therefore,
$\left[ E \right]=\dfrac{\left[ ML{{T}^{-2}} \right]}{\left[ AT \right]}$
$\left[ E \right]=\left[ ML{{A}^{-1}}{{T}^{-3}} \right]$ ……………………… (5)
Now let us find dimension that is asked in the question,
Here again we can exclude the dimensionless constant. Therefore,
$\left[ \dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}} \right]=\left[ {{\varepsilon }_{0}} \right]{{\left[ E \right]}^{2}}$
$\left[ {{\varepsilon }_{0}} \right]{{\left[ E \right]}^{2}}=\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]{{\left[ ML{{A}^{-1}}{{T}^{-3}} \right]}^{2}}$
$\left[ \dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}} \right]=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$

So, the correct answer is “Option C”.

Note: We could avoid this lengthy derivation and solve the question in three simple steps. But for that we have to know that the given quantity is nothing but the energy density of the electric field which is given by energy stored per unit volume. If we ‘u’ represent the energy density given in the question then,
$\left[ u \right]=\dfrac{\left[ energy \right]}{\left[ volume \right]}=\dfrac{\left[ F \right]\left[ L \right]}{{{\left[ L \right]}^{3}}}=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$