Question

solve this physics question on electric field -the center-to-center separation of two tiny, spherical water drops, with identical charges of -1 × 10^-16C is 1cm, calculate the magnitude of the electrostatic force acting between them and the excess electrons on each drop, giving it its charge imbalance?

Answer 1

Question (a)

To calculate the electrostatic force, Coulomb's Law is applied.

$F_{E}=K\frac{q_{1}q_{2}}{r^{2}}$

Where:

• The coulomb constant is $k=9*10^{9}Nw\frac{m^{2}}{C^{2}}$
• The magnitude of electric charges $q_{1}=q_{2}=-1*10^{-16}C$
• The distance between charges is $r=1cm*\frac{1m}{100cm}=0.01m$

Numerically evaluating

$F_{E}=9*10^{9}Nw\frac{m^{2}}{C^{2}}*\frac{-1*10^{-16}C*-1*10^{-16}C}{(0.01m)^{2}}\\ F_{E}=9*10^{-19}Nw$

The electrostatic force between the two drop is

$\boxed{F_{E}=9*10^{-19}Nw}$

Question (b)

The charge of an electron is:$e^{-}=-1.60*10^{-19}C$

The net charge in each drop is $q=-1*10^{-16}C$

Therefore the number of electrons in each sphere is

$N_{e^{-}}=\frac{-1*10^{-16}C}{-1.60*10^{-19} C}=625electrons$

The number of electrons in each drop is.

$\boxed{N_{e^{-}}=625electrons}$