If the capacitance of a nanocapacitor is meas | vedclass

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Question

The unit of capacitance is $\frac {Coulomb}{V} = \frac {Coulomb}{	ext{work done per unit charge}}$=$\frac{	ext { Coulomb }}{\frac{ W }{	ext { Coulo

Vedclass · Accepted Answer

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

Vedclass · Answer

The unit of capacitance is $\frac {Coulomb}{V} = \frac {Coulomb}{	ext{work done per unit charge}}$=$\frac{	ext { Coulomb }}{\frac{ W }{	ext { Coulomb }}}=\frac{	ext { Coulomb }^{2}}{ kgm ^{2} s ^{-2}}= C ^{2} kg ^{-1} m ^{-2} s ^{2}$

Checking units of each option:

$\mu=\frac{ e ^{2} c }{	ext { ha }} \Rightarrow=\frac{	ext { Coulomb }^{2} ms ^{-1}}{ kgm ^{2} s ^{-1} m }= C ^{2} kg ^{-1} m ^{-2} s ^{0}$

$\mu=\frac{ e ^{2} h }{ ca _{\circ}} \Rightarrow=\frac{	ext { Coulomb }^{2} kgm ^{2} s ^{-1}}{ ms ^{-1} m }= C ^{2} kg ^{1} m ^{0} s ^{0}$

$\mu=\frac{	ext { hc }}{ e ^{2} a _{\circ}} \Rightarrow=\frac{ kgm ^{2} s ^{-1} ms ^{-1}}{	ext { Coulomb }^{2} m }= C ^{-2} kg ^{1} m ^{1} s ^{-2}$

$\mu=\frac{ e ^{2} a _{\circ}}{ hc } \Rightarrow=\frac{	ext { Coulomb }^{2} m }{ kgm ^{2} s ^{-1} ms ^{-1}}= C ^{2} kg ^{-1} m ^{-2} s ^{-2}$

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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