If we use permittivity varepsilon , resistan | vedclass

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Question

$\epsilon^{-1}=L^{3} M T^{-2} Q^{-2}, R=L^{2} M T^{-1} Q^{-2}, V=M L^{2} T^{-2} Q^{-1}, G=L^{3} M^{-1} T^{-2}$

As the angular displacement ( $	het

Vedclass · Accepted Answer

all of the above

Vedclass · Answer

$\epsilon^{-1}=L^{3} M T^{-2} Q^{-2}, R=L^{2} M T^{-1} Q^{-2}, V=M L^{2} T^{-2} Q^{-1}, G=L^{3} M^{-1} T^{-2}$

As the angular displacement ( $	heta$ ) is dimensionless so

[displacement] $=\epsilon^{0} R^{0} G^{0} V^{0}$

Velocity, $[v]=L T^{-1}$

here $\epsilon^{-1} R^{-1} G^{0} V^{0}=L^{3} M T^{-2} Q^{-2} L^{-2} M^{-1} T^{1} Q^{2}=L T^{-1}=[v]$

dipole moment, $[p]=Q L$

here $\epsilon^{1} R^{0} G^{0} V^{1}=\left(L^{-3} M^{-1} T^{2} Q^{2}ight)\left(M L^{2} T^{-2} Q^{-1}ight)=Q L^{-1} 
eq[p]$

force, $[f]=\epsilon^{-1} L^{-2} Q^{2}$

here $\epsilon^{1} R^{0} G^{0} V^{2}=\left(L^{-3} M^{-1} T^{2} Q^{2}ight)\left(M^{2} L^{4} T^{-4} Q^{-2}ight)=\left(L^{3} M T^{-2} Q^{-2}ight) L^{-2} Q^{2}=$

$\epsilon^{-1} L^{-2} Q^{2}=[f]$

If we use permittivity $ \varepsilon $, resistance $R$, gravitational constant $G$ and voltage $V$ as fundamental physical quantities, then

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