
The dimensional formula for capacitance is ________. Take \[Q\] as the dimension formula of charge-
(A). \[{{M}^{1}}{{L}^{-2}}{{T}^{-2}}{{Q}^{-2}}\]
(B). \[{{M}^{1}}{{L}^{2}}{{T}^{-2}}{{Q}^{-2}}\]
(C). \[{{M}^{1}}{{L}^{-2}}{{T}^{2}}{{Q}^{2}}\]
(D). \[{{M}^{-1}}{{L}^{-2}}{{T}^{2}}{{Q}^{2}}\]
Answer
485.7k+ views
Hint: Capacitance is the ability of a conductor to store charge on it. It is the ratio of charge on a conductor to the potential drop on it. Its Dimensional formula is represented in terms of fundamental units. Voltage is the work done in an electric field while charge is also assumed as a fundamental unit.
Formula used: \[C=\dfrac{Q}{V}\]
\[V=\,E\cdot d\]
\[E=\dfrac{F}{q}\]
Complete step by step answer:
Dimensional formula is the representation of a physical parameter in terms of the fundamental units.
Capacitance is defined as the ability of a conductor to store charge inside it. Its SI unit is farad ( \[F\] ). It is given by-
\[C=\dfrac{Q}{V}\] ---- (1)
Here, \[C\] is the capacitance
\[Q\] is the charge on the conductor
\[V\] is the magnitude of potential drop
As we know that,
\[V=\,E\cdot d\]
Here, \[E\] is the electric field
\[d\] is the distance between the plates of the capacitor
As \[E=\dfrac{F}{q}\] ( \[F\] is the electrostatic force), the dimensional formula of \[E\] is-
\[\begin{align}
& [E]=\dfrac{[ML{{T}^{-2}}]}{[Q]} \\
& [E]=[ML{{T}^{-2}}{{Q}^{-1}}] \\
\end{align}\]
Therefore, the dimensional formula for \[V\] is-
\[[V]=[ML{{T}^{-2}}{{Q}^{-1}}][L]\]
\[\therefore [V]=[M{{L}^{2}}{{T}^{-2}}{{Q}^{-1}}]\] ----- (2)
From eq (1) and eq (2), the dimensional formula for capacitance will be-
\[\begin{align}
& [C]=\dfrac{[Q]}{[M{{L}^{2}}{{T}^{-2}}{{Q}^{-1}}]} \\
& \Rightarrow [C]=[{{M}^{-1}}{{L}^{-2}}{{T}^{2}}{{Q}^{2}}] \\
\end{align}\]
Therefore, the dimensional formula of capacitance is \[[{{M}^{-1}}{{L}^{-2}}{{T}^{2}}{{Q}^{2}}]\] .
So, the correct answer is “Option D”.
Note: Coulomb’s law states that force due to an electrostatic charge is directly proportional to the product of magnitude of charges and inversely proportional to the square of distance between them. Voltage is the work done to bring a unit charge from infinity to the electric field of a charge. The dimensional formula for \[Q\] is \[[IT]\] .
Formula used: \[C=\dfrac{Q}{V}\]
\[V=\,E\cdot d\]
\[E=\dfrac{F}{q}\]
Complete step by step answer:
Dimensional formula is the representation of a physical parameter in terms of the fundamental units.
Capacitance is defined as the ability of a conductor to store charge inside it. Its SI unit is farad ( \[F\] ). It is given by-
\[C=\dfrac{Q}{V}\] ---- (1)
Here, \[C\] is the capacitance
\[Q\] is the charge on the conductor
\[V\] is the magnitude of potential drop
As we know that,
\[V=\,E\cdot d\]
Here, \[E\] is the electric field
\[d\] is the distance between the plates of the capacitor
As \[E=\dfrac{F}{q}\] ( \[F\] is the electrostatic force), the dimensional formula of \[E\] is-
\[\begin{align}
& [E]=\dfrac{[ML{{T}^{-2}}]}{[Q]} \\
& [E]=[ML{{T}^{-2}}{{Q}^{-1}}] \\
\end{align}\]
Therefore, the dimensional formula for \[V\] is-
\[[V]=[ML{{T}^{-2}}{{Q}^{-1}}][L]\]
\[\therefore [V]=[M{{L}^{2}}{{T}^{-2}}{{Q}^{-1}}]\] ----- (2)
From eq (1) and eq (2), the dimensional formula for capacitance will be-
\[\begin{align}
& [C]=\dfrac{[Q]}{[M{{L}^{2}}{{T}^{-2}}{{Q}^{-1}}]} \\
& \Rightarrow [C]=[{{M}^{-1}}{{L}^{-2}}{{T}^{2}}{{Q}^{2}}] \\
\end{align}\]
Therefore, the dimensional formula of capacitance is \[[{{M}^{-1}}{{L}^{-2}}{{T}^{2}}{{Q}^{2}}]\] .
So, the correct answer is “Option D”.
Note: Coulomb’s law states that force due to an electrostatic charge is directly proportional to the product of magnitude of charges and inversely proportional to the square of distance between them. Voltage is the work done to bring a unit charge from infinity to the electric field of a charge. The dimensional formula for \[Q\] is \[[IT]\] .
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