E.m.f has the same unit as electric potential, hence they are dimensionally equivalent. From Electrostatics, potential difference = kq/r where k=constant (Nm2C-2), q=charge (C), r=distance (m). Find the dimensions of P.d.

$$\mathrm{Pd} = \frac{k q}{r}$$

where $[k] = \mathrm{N \cdot m^2 \cdot C^{-2}}$, $[q] = \mathrm{C}$, $[r] = \mathrm{m}$.

Substitute dimensions:

$$[\mathrm{Pd}] = \frac{[\mathrm{N \cdot m^2 \cdot C^{-2}}] \cdot [\mathrm{C}]}{[\mathrm{m}]} = \mathrm{N \cdot m \cdot C^{-1}}.$$

$$[\mathrm{N}] = [\mathrm{F}] = \mathrm{M L T^{-2}}.$$

Thus,

$$[\mathrm{Pd}] = (\mathrm{M L T^{-2}}) \cdot \mathrm{L} \cdot \mathrm{C^{-1}} = \mathrm{M L^2 T^{-2} C^{-1}}.$$

$$[\mathrm{C}] = [\mathrm{A \cdot s}] = \mathrm{I T}.$$

Substitute:

$$[\mathrm{Pd}] = \mathrm{M L^2 T^{-2}} \cdot (\mathrm{I T})^{-1} = \mathrm{M L^2 T^{-2} \cdot I^{-1} T^{-1}} = \mathrm{M L^2 T^{-3} I^{-1}}.$$

Work done $= \mathrm{F \cdot r}$, so

$$[\text{work}] = (\mathrm{M L T^{-2}}) \cdot \mathrm{L} = \mathrm{M L^2 T^{-2}}.$$ $$[\mathrm{Pd}] = \frac{[\text{work}]}{[q]} = \frac{\mathrm{M L^2 T^{-2}}}{\mathrm{I T}} = \mathrm{M L^2 T^{-3} I^{-1}}.$$

has same dimensions as electric potential difference.

$$\mathrm{[Pd]} = \mathrm{M L^{2} T^{-3} I^{-1}}$$