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Question

Physics Question on physical world

Dimensions of capacitance is.

A

[M−1L−2T4A2][{{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{A}^{2}}]

B

[MLT−3A−1][ML{{T}^{-3}}{{A}^{-1}}]

C

[ML2T−3A−1][M{{L}^{2}}{{T}^{-3}}{{A}^{-1}}]

D

[M−1L−2T3A−1][{{M}^{-1}}{{L}^{-2}}{{T}^{3}}{{A}^{-1}}]

Answer

[M−1L−2T4A2][{{M}^{-1}}{{L}^{-2}}{{T}^{4}}{{A}^{2}}]

Explanation

Solution

The capacitance (C)(C) of a conductor is defined as the ratio of charge (q)(q) given to the rise in potential (V)(V) of the conductor. That is C=qVC=\frac{q}{V} ∴ Farad = coulomb  volt = coulomb  joule / coulomb \therefore \text { Farad }=\frac{\text { coulomb }}{\text { volt }}=\frac{\text { coulomb }}{\text { joule / coulomb }} = coulomb 2 joule =\frac{\text { coulomb }^{2}}{\text { joule }} =( ampere-sec )2 newton-metre = ampere 2−sec⁡2(kg−msec−2)× metre =\frac{(\text { ampere-sec })^{2}}{\text { newton-metre }}=\frac{\text { ampere }^{2}-\sec ^{2}}{\left( kg - m sec ^{-2}\right) \times \text { metre }} = ampere 2−sec4kg−metre2=\frac{\text { ampere }^{2}- sec ^{4}}{ kg - metre ^{2}} =kg−1−metre−2−sec⁡4−ampere2= kg ^{-1}- metre ^{-2}-\sec ^{4}- ampere ^{2} Hence, dimensions of capacitance are [M−1L−2T4A2]\left[ M ^{-1} L ^{-2} T ^{4} A ^{2}\right]