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The dimension of $\dfrac{1}{2}{\varepsilon _0}{E^2}$ ( ${\varepsilon _0}$ is the permittivity of free space; E is the electric field) is?
A) $ML{T^{ - 2}}$
B) $M{L^2}{T^{ - 2}}$
C) $M{L^{ - 1}}{T^{ - 2}}$
D) $M{L^2}{T^{ - 1}}$

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Last updated date: 14th Jun 2024
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Answer
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Hint: This $\dfrac{1}{2}{\varepsilon _0}{E^2}$is the equation that represents the energy density of the capacitor. It represents the energy per unit volume. The dimensional formula for the $\dfrac{1}{2}{\varepsilon _0}{E^2}$ can be found out by writing every term in the equation in its corresponding dimensions.

Complete step by step solution:
We are interested to find out the dimension of $\dfrac{1}{2}{\varepsilon _0}{E^2}$
From the electrostatic force of attraction relation, we know that,
${F_e} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
However, we are interested in it${\varepsilon _0}$. Thus on rearranging the equation we get,
${\varepsilon _0} = \dfrac{{{q^2}}}{{4\pi {F_e}{r^2}}}$
Let's write all the quantities in the equation in its corresponding dimension.
Dimension for a charge, [$\text{current}$ $\times$ $\text{time}$] $= [AT]$
Dimension for force ${F_e}$ is equal to ${F_e} = ma$,
=${\text{[M][L}}{{\text{T}}^{{\text{ - 2}}}}{\text{]}}$ where, ${\text{[M]}}$=dimension of the mass of an object, ${\text{[L}}{{\text{T}}^{{\text{ - 2}}}}{\text{]}}$=dimension of acceleration of an object
The dimension of distance, ${{\text{r}}^{\text{2}}}{\text{ = }}\left[ {{{\text{L}}^{\text{2}}}} \right]$
Put all the dimension in the rearranged equation we get,
${\varepsilon _0} = \dfrac{{{{\left[ {AT} \right]}^2}}}{{\left[ {ML{T^{ - 2}}} \right]{{\left[ L \right]}^2}}}$
Since $4\pi $ is a dimensionless quantity.it can be neglected from the dimension formula.
Then, on simplification we get,
${\varepsilon _0} = \dfrac{{\left[ {{A^2}{T^2}} \right]}}{{\left[ {ML{T^{ - 2}}} \right]{{\left[ L \right]}^2}}} = \dfrac{{\left[ {{A^2}{T^2}} \right]}}{{\left[ {M{L^3}{T^{ - 2}}} \right]}}$
$\Rightarrow \dfrac{{\left[ {{A^2}{T^2}} \right]}}{{\left[ {M{L^3}{T^{ - 2}}} \right]}} = \left[ {{A^2}{T^2}} \right]{\left[ {M{L^3}{T^{ - 2}}} \right]^{ - 1}}$
$\Rightarrow {\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$
Similarly, the electric field is given by,
$E = \dfrac{F}{q}$
On applying the dimensional formula for the electric field equation we get,
$\Rightarrow E = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {AT} \right]}}$
On simplification we get,
$\Rightarrow E = \left[ {ML{T^{ - 3}}{A^{ - 1}}} \right]$
Now let's consider the equation which is given $\dfrac{1}{2}{\varepsilon _0}{E^2}$
We know that Here ½ is a dimensionless quantity.
Let’s put the dimensions of ${\varepsilon _0}$ and E in the equation we get,
$\Rightarrow \dfrac{1}{2}{\varepsilon _0}{E^2}$$ = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right] \times {\left[ {ML{T^{ - 3}}{A^{ - 1}}} \right]^2}$
On simplifying the above terms, we get
$ = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right] \times \left[ {{M^2}{L^2}{T^{ - 6}}{A^{ - 2}}} \right]$
$ = \left[ {M{L^1}{T^{ - 2}}} \right]$

$\therefore$ The dimension of $\dfrac{1}{2}{\varepsilon _0}{E^2}$ is $M{L^{ - 1}}{T^{ - 2}}$. The correct option is (C).

Additional information:
The dimensions of a physical quantity are defined as the powers of fundamental quantities to represent physical quantity. The expression showing the powers to which the fundamental quantities must be raised to represent a physical quantity is called dimensional formula.
Dimensions for the fundamental physical quantities.

1) Mass[ M ]
2) Length[ L ]
3) Time[ T ]
4) Temperature[ K ]
5) Electric current[ A ]
6) Luminous intensity[ cd ]
7) Amount of substance[ mol ]


Constants having dimensional formulas are called Dimensional constants.
e.g.: Planck’s constant, speed of light, Universal gravitational constant.
Physical quantities having no dimensional formula are called Dimensionless quantities.
e.g.: Angle, Strain, and Relative Density.
Limitations of Dimensional analysis.
Proportionality constants cannot be determined by dimensional analysis.
Formulae containing non- algebraic functions like sin, cos, log, exponential, etc., cannot be derived.
Dimensional does not bifurcate between a scalar and a vector quantity.

Note:
Square bracket [ ] is used to represent the dimension of the physical quantity
Applications of Dimensional analysis.
Dimensional formulae can be used to convert one system of the unit to another system of the unit.
Dimensional formulae can be used to check the correctness of a given equation.
Dimensional formulae can be used to derive the relationship among different physical quantities.