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Asked in JEE Main 22nd Jan 2nd Shift 2025 · Capacitance of a conductor and a capacitor
C=Q/V, so the dimensions of capacitance are those of charge divided by those of potential.
Potential is work per unit charge: [V]=([ML²T⁻²])/([C])=[M L²T⁻²C⁻¹].
Hence [C_cap]=([C])/([ML²T⁻²C⁻¹])=[C²M⁻¹L⁻²T²].
So the farad has dimensions [C²M⁻¹L⁻²T²].
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