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If we use permittivity $ \varepsilon $, resistance $R$, gravitational constant $G$ and voltage $V$ as fundamental physical quantities, then
[angular displacement] $= \varepsilon^0R^0G^0V^0$
[Velocity] = $\varepsilon ^{-1}R^{-1}G^0V^0$
[force] = $ \varepsilon ^1R^0G^0V^2$
all of the above
Solution
$\epsilon^{-1}=L^{3} M T^{-2} Q^{-2}, R=L^{2} M T^{-1} Q^{-2}, V=M L^{2} T^{-2} Q^{-1}, G=L^{3} M^{-1} T^{-2}$
As the angular displacement ( $\theta$ ) 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 $\epsilon^{1} R^{0} G^{0} V^{1}=\left(L^{-3} M^{-1} T^{2} Q^{2}\right)\left(M L^{2} T^{-2} Q^{-1}\right)=Q L^{-1} \neq[p]$
force, $[f]=\epsilon^{-1} L^{-2} Q^{2}$
here $\epsilon^{1} R^{0} G^{0} V^{2}=\left(L^{-3} M^{-1} T^{2} Q^{2}\right)\left(M^{2} L^{4} T^{-4} Q^{-2}\right)=\left(L^{3} M T^{-2} Q^{-2}\right) L^{-2} Q^{2}=$
$\epsilon^{-1} L^{-2} Q^{2}=[f]$
