If $$C$$ and $$V$$ represent capacity and voltage respectively then what are the dimensions of $$\lambda$$ where $$C/V = \lambda$$?

We need to find the dimensions of $$\lambda = \frac{C}{V}$$, where $$C$$ is capacitance and $$V$$ is voltage.

The dimensions of capacitance are $$[C] = [M^{-1}L^{-2}I^2T^4]$$. This follows from $$C = \frac{Q}{V} = \frac{IT}{V}$$ and $$V = \frac{W}{Q} = \frac{ML^2T^{-2}}{IT} = [ML^2I^{-1}T^{-3}]$$, giving $$[C] = \frac{[IT]}{[ML^2I^{-1}T^{-3}]} = [M^{-1}L^{-2}I^2T^4]$$.

The dimensions of voltage are $$[V] = [ML^2I^{-1}T^{-3}]$$.

Therefore, the dimensions of $$\lambda = \frac{C}{V}$$ are $$[\lambda] = \frac{[M^{-1}L^{-2}I^2T^4]}{[ML^2I^{-1}T^{-3}]} = [M^{-1-1}L^{-2-2}I^{2+1}T^{4+3}] = [M^{-2}L^{-4}I^3T^7]$$.