Question

Two tiny, spherical water drops, with identical charges of -4.29 Ã 10-16 C, have a center-to-center separation of 0.845 cm. A) What is the magnitude of the electrostatic force acting between them? B) How many excess electrons are on each drop, giving it its charge imbalance?

Answer 1

Answer:

Electrostatic force, $F=2.31\times 10^{-17}\ N$

Number of electrons, n = 2681 electrons

Explanation:

Given that,

Charges, $q_=q_2=-4.29\times 10^{-16}\ C$

Separation between charges, $r=0.845\ cm=0.845\times 10^{-2}\ m$

(a) Let F is the magnitude of the electrostatic force acting between them. The electric force between charges is given by :

$F=\dfrac{kq^2}{r^2}$

$F=\dfrac{9\times 10^9\times (4.29\times 10^{-16})^2}{(0.845\times 10^{-2})^2}$

$F=2.31\times 10^{-17}\ N$

(b) Let n be the excess electrons on each drop, giving it its charge imbalance. It can be calculated using quantization of electric charge as :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{4.29\times 10^{-16}}{1.6\times 10^{-19}}$

n = 2681.25 electrons

or

n = 2681 electrons

So, the number of excess electrons on each drop is 2681 electrons. Hence, this is the required solution.