A vertical electric field of magnitude $$4.9 \times 10^{5}$$ N C$$^{-1}$$ just prevents a water droplet of a mass $$0.1$$ g from falling. The value of charge on the droplet will be : (Given $$g = 9.8$$ m s$$^{-2}$$)

We are given: electric field $$E = 4.9 \times 10^5$$ N/C (vertical, upward), mass of water droplet $$m = 0.1$$ g $$= 0.1 \times 10^{-3}$$ kg $$= 10^{-4}$$ kg, $$g = 9.8$$ m/s$$^2$$. The electric force must balance the gravitational force to prevent the droplet from falling:

$$ qE = mg $$

Solving for the charge $$q$$, we get:

$$ q = \frac{mg}{E} = \frac{10^{-4} \times 9.8}{4.9 \times 10^5} $$

$$ q = \frac{9.8 \times 10^{-4}}{4.9 \times 10^5} $$

$$ q = 2 \times 10^{-9} \text{ C} $$

Therefore, the correct answer is Option B.