Solveeit Logo

Question

Question: What is the dimensional formula of farad? A) \(\left[ {{M^{ - 1}}{L^{ - 2}}TQ} \right]\) B) \(\l...

What is the dimensional formula of farad?
A) [M1L2TQ]\left[ {{M^{ - 1}}{L^{ - 2}}TQ} \right]
B) [M1L2T2Q2]\left[ {{M^{ - 1}}{L^{ - 2}}{T^2}{Q^2}} \right]
C) [M1L2TQ2]\left[ {{M^{ - 1}}{L^{ - 2}}T{Q^2}} \right]
D) [M1L2T2Q]\left[ {{M^{ - 1}}{L^{ - 2}}{T^2}Q} \right]

Explanation

Solution

To solve this question we should know about the base quantities which are used to form the dimensional formulae of any quantity. Also we should know how farad is calculated i.e., the quantities involved in its calculation and their dimensional formulae.

Formulae used:
C=qVC = \dfrac{q}{V}
Here CC is the capacitance of the capacitor, qq is the charge stored in it and VV is the potential difference across the capacitor.
V=WqV = \dfrac{W}{q}
Here VV is the potential difference across the capacitor, WW is the work done by the charge and qq is the charge stored in it.

Complete step by step answer:
To solve this question we should know what farad is. One Farad can be defined as the capacitance of a capacitor when the charge stored is of one coulomb and the potential difference across the capacitor is of one volt.
So,
C=qV\Rightarrow C = \dfrac{q}{V}
Here CC is the capacitance of the capacitor, qq is the charge stored in it and VV is the potential difference across the capacitor.
Let this be equation 1.
1Farad=1coulomb1Volt\Rightarrow 1 Farad = \dfrac{{1 coulomb}}{{1 Volt}}
So the dimensional formula of farad and coulomb will be the same.
We know that,
V=WqV = \dfrac{W}{q}
Here VV is the potential difference across the capacitor, WW is the work done by the charge and qq is the charge stored in it.
Let this be equation 2.
Substituting the values of equation 2 in equation 1 we get,
C=q2W\Rightarrow C = \dfrac{{{q^2}}}{W}
Let this be equation 3.
We know that the dimensional formulae of
[q]=[Q]\left[ q \right] = \left[ Q \right] and
[W]=[ML2T2]\left[ W \right] = \left[ {M{L^2}{T^{ - 2}}} \right]
Substituting the values of the above quantities in the equation 3 we get,
[C]=[Q]2[ML2T2]\Rightarrow \left[ C \right] = \dfrac{{{{\left[ Q \right]}^2}}}{{\left[ {M{L^2}{T^{ - 2}}} \right]}}
[C]=[M1L2T2Q2]\Rightarrow \left[ C \right] = \left[ {{M^{ - 1}}{L^{ - 2}}{T^2}{Q^2}} \right]

So the answer will be option (B).

Note: To solve questions related to dimensional analysis of any quantity, break the quantity into its smaller known units. Use the dimensional formulae of the smaller known units to find the dimensional formulae of the given quantity.