Question

Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +18 pC, fairly typical values.

Answer 1

To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is

$V = \frac{kq}{r}$

k = Coulomb's constant

q = Charge

r = Distance between them

$q = 18 pC \rightarrow q = 1.8*10^-11 C$

$d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m$

Replacing,

$V = \frac{kq}{r}$

$V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}$

$V = 135 V$

Therefore the potential at the surface of the raindrop is 135 V