If the capacitance of a nanocapacitor is meas | vedclass

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Question

(D) The dimension of capacitance $C$ is $[M^{-1} L^{-2} T^4 I^2]$.We analyze the dimensions of the given constants:Charge $e = [I T]$Bohr radius $a_0

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$u = \frac{e^2 a_0}{hc}$

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(D) The dimension of capacitance $C$ is $[M^{-1} L^{-2} T^4 I^2]$.We analyze the dimensions of the given constants:Charge $e = [I T]$Bohr radius $a_0 = [L]$Planck's constant $h = [M L^2 T^{-1}]$Speed of light $c = [L T^{-1}]$Now,check the dimensions of option $C$ $(u = \frac{e^2 c}{h a_0})$:Dimensions of $\frac{e^2 c}{h a_0} = \frac{[I T]^2 [L T^{-1}]}{[M L^2 T^{-1}] [L]} = \frac{[I^2 T^2 L T^{-1}]}{[M L^3 T^{-1}]} = [M^{-1} L^{-2} T^2 I^2]$.Wait,let us re-evaluate the dimensions of capacitance. Capacitance $C = \frac{Q}{V} = \frac{Q}{W/Q} = \frac{Q^2}{W} = \frac{[I^2 T^2]}{[M L^2 T^{-2}]} = [M^{-1} L^{-2} T^4 I^2]$.Checking option $D$ $(u = \frac{e^2 a_0}{hc})$:Dimensions of $\frac{e^2 a_0}{hc} = \frac{[I^2 T^2] [L]}{[M L^2 T^{-1}] [L T^{-1}]} = \frac{[I^2 T^2 L]}{[M L^3 T^{-2}]} = [M^{-1} L^{-2} T^4 I^2]$.Since the dimensions of $u = \frac{e^2 a_0}{hc}$ match the dimensions of capacitance,option $D$ is correct.

If the capacitance of a nanocapacitor is measured in terms of a unit $u$ made by combining the electric charge $e,$ Bohr radius $a_0,$ Planck's constant $h$ and speed of light $c,$ then

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