Physics

A quantity $$X$$ is given by $$\epsilon_0 L\dfrac {\Delta V}{\Delta t}$$, where $$\epsilon_0$$ is the permittivity of fee space, $$L$$ is length, $$\Delta V$$ is a potential difference and $$\Delta t$$ is a time interval. The dimensional formula for $$X$$ is the same as that of


ANSWER

current


SOLUTION
Dimensional formula for $$ \epsilon_0 = [M^{-1} L^{-3} T^{4} I^{2}]$$
Dimensional formula for $$ L = [L]$$
Dimensional formula for $$ \Delta V = [M L^2 T^{-3} I^{-1}]$$
Dimensional formula for $$ \Delta t = [T^{1}]$$
Therefore, dimensional formula for $$ X $$ is: $$ \dfrac{ [M^{-1} L^{-3} T^{4} I^{2}] [L] [M L^2 T^{-3} I^{-1}] }{ [T^{1}] }=[I] $$
Therefore, its the same as that of current.
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