Question

Identify the dimension of electric potential.
$
  {\text{A}}{\text{. }}{L^2}M{T^{ - 3}}{I^{ - 1}} \\
  {\text{B}}{\text{. }}{L^1}M{T^{ - 2}}{I^{ - 1}} \\
  {\text{C}}{\text{. }}{L^{ - 2}}M{T^{ - 2}}{I^{ - 1}} \\
  {\text{D}}{\text{. none of the above}} \\
$

Answer 1

Hint: The dimensions of a physical quantity can be found out from the relation of that physical quantity with the fundamental quantities. We know that electrical potential energy is equal to the product of charge and its electric potential.

Detailed step by step answer:
Dimensional formula: A dimensional formula of a physical quantity is an expression which describes the dependence of that quantity on the fundamental quantities.
All physical quantities can be expressed in terms of certain fundamental quantities. Following table contains the fundamental quantities and their units and dimensional notation respectively.

No.	Quantities	unit	Dimensional formula
1.	Length	metre (m)	$\left[ {{M^0}{L^1}{T^0}} \right]$
2.	Mass	kilogram (g)	$\left[ {{M^1}{L^0}{T^0}} \right]$
3.	Time	second (s)	$\left[ {{M^0}{L^0}{T^1}} \right]$
4.	Electric current	ampere (A)	$\left[ {{M^0}{L^0}{T^0}{A^1}} \right]$
5.	Temperature	kelvin [K]	$\left[ {{M^0}{L^0}{T^0}{K^1}} \right]$
6.	Amount of substance	mole [mol]	$\left[ {{M^0}{L^0}{T^0}mo{l^1}} \right]$
7.	Luminous intensity	candela [cd]	$\left[ {{M^0}{L^0}{T^0}C{d^1}} \right]$

We need to find out the dimensional formula for electric potential (V). It can be calculated using the relation of electric potential with potential energy.
As we know that electrical potential energy is equal to product of charge and its electric potential, we can write
$E = qV$
[Potential energy] = [charge]$ \times $[electric potential]
$ \Rightarrow $ [Electric potential] = [Potential energy] $ \times $[charge]$^{ - 1}$
Potential energy is just another form of energy, so, its dimensions are same as that of energy which are given as
[Potential energy]$ = \left[ {{M^1}{L^2}{T^{ - 2}}} \right]$
Also we know that electric charge is equal to the product of current and time.
$q = It$
$\therefore $[charge]$ = \left[ {{M^0}{L^0}{T^1}{I^1}} \right]$
$ \Rightarrow $[charge]$^{ - 1} = \left[ {{M^0}{L^0}{T^{ - 1}}{I^{ - 1}}} \right]$
$\therefore $ [Electric potential]$ = \left[ {{M^1}{L^2}{T^{ - 2}}} \right] \times \left[ {{M^0}{L^0}{T^{ - 1}}{I^{ - 1}}} \right] = \left[ {{M^1}{L^2}{T^{ - 3}}{I^{ - 1}}} \right]$
Hence, the correct answer is option A.

Note: Mass, length and time are most commonly encountered fundamental quantities so they must be specified in all dimensional formulas. The square bracket notation is used only for dimensional formulas.