Electric field in a certain region is given by $$\vec{E} = \frac{A}{x^2}\hat{i} + \frac{B}{y^3}\hat{j}$$. The SI unit of $$A$$ and $$B$$ are:

We need to find the SI units of $$A$$ and $$B$$ in the electric field expression $$\vec{E} = \frac{A}{x^2}\hat{i} + \frac{B}{y^3}\hat{j}$$.

Finding the unit of A:

The term $$\frac{A}{x^2}$$ represents the x-component of electric field.

$$[A] = [E] \times [x^2] = \text{N C}^{-1} \times \text{m}^2 = \text{N m}^2 \text{ C}^{-1}$$

Finding the unit of B:

The term $$\frac{B}{y^3}$$ represents the y-component of electric field.

$$[B] = [E] \times [y^3] = \text{N C}^{-1} \times \text{m}^3 = \text{N m}^3 \text{ C}^{-1}$$

Therefore, $$A$$ has SI unit N m$$^2$$ C$$^{-1}$$ and $$B$$ has SI unit N m$$^3$$ C$$^{-1}$$.

The correct answer is Option 2.