If $$\epsilon$$, E and t represent the free space permittivity, electric field and time respectively, then the unit of $$ \frac{\epsilon E}{t}$$ will be :

The electric displacement vector $$\mathbf{D}$$ is related to the electric field $$\mathbf{E}$$ by the constitutive relation $$\mathbf{D} = \epsilon \mathbf{E}$$.

By definition, the surface integral $$\iint \mathbf{D}\,\cdot d\mathbf{S}$$ equals the free charge $$q$$ enclosed. Hence the unit of $$\mathbf{D}$$ (and therefore of $$\epsilon E$$) must be the unit of charge per unit area, that is $$\text{coulomb per square metre}$$, written as $$C/m^{2}$$.

We now divide by time $$t$$, whose SI unit is the second, $$s$$:

$$\frac{\epsilon E}{t} \; \text{has unit} \; \frac{C/m^{2}}{s} = \frac{C}{m^{2}\,s}.$$

Since $$1\,\text{ampere} = 1\,\text{coulomb}\,\text{s}^{-1}$$, we replace $$C/s$$ by $$A$$:

$$\frac{C}{m^{2}\,s} = \frac{A}{m^{2}}.$$

Thus, the unit of $$\dfrac{\epsilon E}{t}$$ is $$A/m^{2}$$.

Correct option: Option B.