Question

If${\varepsilon _0}$is the permittivity of free space and$E$is the electric field, then the dimensions of$\dfrac{1}{2}{\varepsilon _0}{E^2}$is
A. $\left[ {MLT} \right]$
B. $\left[ {M{L^2}{T^{ - 2}}} \right]$
C. $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
D. $\left[ {M{L^2}{T^{ - 1}}} \right]$

Answer 1

Electric field is given as force per unit charge. Therefore it will have the dimensional formula of force $\left[ {ML{T^{ - 2}}} \right]$ and charge $\left[ {IT} \right]$. The permittivity of free space represents the energy that is stored in an electric field and has the dimensional formula as ${\varepsilon _0} = \left[ {{M^{ - 1}}\;{L^{ - 3}}\;{T^{4\;}}{I^2}} \right]$.

Formula Used:
The dimensional formula for electric field is:$E = \left[ {ML{T^{ - 3}}{I^{ - 1}}} \right]$
The dimensional formula for permittivity of free space is given by:${\varepsilon _0} = \left[ {{M^{ - 1}}\;{L^{ - 3}}\;{T^{4\;}}{I^2}} \right]$

Complete step by step answer:
Dimensions of a physical quantity are the powers to which the basic units should be raised to represent the unit of that physical quantity. Dimensional formula of a physical quantity is the mathematical expression which provides information about the quantities that are involved in the unit of that physical quantity and also the powers to which they should be raised.
$\left[ M \right]$stands for the dimensional formula for mass. Similarly, $\left[ L \right]$stands for length and $\left[ T \right]$ stands for time.
The formula for electric field is Force per unit charge. The dimensional formula for force is $\left[ {ML{T^{ - 2}}} \right]$and that for charge is $\left[ {IT} \right]$( if that for current is written as $\left[ I \right]$). Then, the dimensional formula for electric field is:

\Rightarrow E = \left[ {ML{T^{ - 3}}{I^{ - 1}}} \right]$$ $$\Rightarrow{E^2} = \left[ {{M^2}{L^2}{T^{ - 6}}{I^{ - 2}}} \right]$$$$ \to $$(1) On the other hand, the dimensional formula for permittivity of free space is given by $${\varepsilon _0} = \left[ {{M^{ - 1}}\;{L^{ - 3}}\;{T^{4\;}}{I^2}} \right]$$$$ \to $$(2) Therefore, $$\dfrac{1}{2}{\varepsilon _0}{E^2} = \left[ {{M^{ - 1}}\;{L^{ - 3}}\;{T^{4\;}}{I^2}} \right]\left[ {{M^2}{L^2}{T^{ - 6}}{I^{ - 2}}} \right] = \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right]$$ **Hence, option C is the correct answer.** **Note:** A physical quantity may be dimensionless, but it may have units. For example, permittivity of free space is a constant and therefore is a dimensionless quantity. Though it is constant, it has units. However if a quantity has (non-zero) dimensions, it cannot be unit less. Taking the square of the electric field is equal to taking the square of its whole dimensional formula (all the powers of dimensional formula will be multiplied by 2).