A quantity X is given by \(\varepsilon_0L\dfrac{\Delta V}{\Delta t}\), where ε0 is the permittivity of free space, L is the length, ΔV is a potential difference and Δt is a time interval. The dimension of X is same as that of
Current
Dimensional analysis is easiest here through a known formula: the capacitance of a parallel-plate capacitor is \(C=\dfrac{\varepsilon_0A}{d}\), so the product of the permittivity and a length has the dimensions of capacitance.
Step 1 — \([\varepsilon_0L]=[C]\).
Step 2 — Capacitance times potential difference is charge: \([\varepsilon_0L\,\Delta V]=[CV]=[Q]\).
Step 3 — Charge divided by time is current: \([X]=\left[\dfrac{Q}{t}\right]=[I]=\text{A}\).
Step 4 — Check in base dimensions: \([\varepsilon_0]=\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2},\ [V]=\text{M}\text{L}^{2}\text{T}^{-3}\text{A}^{-1}\), so \([X]=\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^{2}\cdot\text{L}\cdot\text{M}\text{L}^{2}\text{T}^{-3}\text{A}^{-1}\cdot\text{T}^{-1}=\text{A}\).
Resistance has dimensions ML2T−3A−2, charge is AT (which X would be only without the division by Δt), and voltage is ML2T−3A−1; none of these equals A.
Current option is correct.
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