If mass (M),length (L),time (T),and electric | vedclass

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(B) According to Coulomb's law,the force between two charges is given by $F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2}$.Rearranging for pe

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$M^{-1} L^{-3} T^4 A^2$

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(B) According to Coulomb's law,the force between two charges is given by $F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2}$.Rearranging for permittivity,we get $\epsilon_0 = \frac{q_1 q_2}{4 \pi F r^2}$.The dimensional formula for charge $(q)$ is $[AT]$,for force $(F)$ is $[MLT^{-2}]$,and for distance $(r)$ is $[L]$.Substituting these dimensions: $\epsilon_0 = \frac{[AT][AT]}{[MLT^{-2}][L^2]} = \frac{[A^2T^2]}{[ML^3T^{-2}]}$.Simplifying the expression,we get $[M^{-1} L^{-3} T^4 A^2]$.

Column-$I$	Column-$II$
$(A)$ Electrical resistance	$(p)$ $M L^3 T^{-3} A^{-2}$
$(B)$ Electrical potential	$(q)$ $M L^2 T^{-3} A^{-2}$
$(C)$ Specific resistance	$(r)$ $M L^2 T^{-3} A^{-1}$
$(D)$ Specific conductance	$(s)$ None of these

If mass $(M)$,length $(L)$,time $(T)$,and electric current $(A)$ are taken as fundamental quantities,then the dimensional formula of permittivity $(\epsilon_0)$ is:

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Match the following two columns:

Column-$I$ Column-$II$

$(A)$ Electrical resistance $(p)$ $M L^3 T^{-3} A^{-2}$

$(B)$ Electrical potential $(q)$ $M L^2 T^{-3} A^{-2}$

$(C)$ Specific resistance $(r)$ $M L^2 T^{-3} A^{-1}$

$(D)$ Specific conductance $(s)$ None of these

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