Question

Two tiny, spherical water drops, with identical charges of -6.19 × 10-16 C, have a center-to-center separation of 1.22 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:

a

  $F = 2.32*10^{-17} \ N$

b

$n =3869 \ electrons$

Explanation:

From the question we are told that

   The  charge on each water drop is  $q_1=q_2=q = - 6.19*10^{-16} \ C$

   The  distance of separation is  $d = 1.22\ cm = 0.0122 \ m$  

   

Generally the electrostatic force between the water drops is mathematically represented as

      $F = \frac{k * q_1 * q_2 }{ d^ 2}$

Here  k is the coulombs constant with value  $k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

So  

      $F = \frac{9*10^9 * -6.19 *10^{-16} * (-6.19*10^{-16}) }{ 0.0122^ 2}$

       $F = 2.32*10^{-17} \ N$

Generally the quantity of charge is mathematically represented as

        $q = n * e$

Here n is the number of electron present

and  e is the charge on one electron with value  $e = 1.60*10^{-19} \ C$

    So

         $n = \frac{6.19 *10^{-16}}{1.60*10^{-19}}$

         $n =3869 \ electrons$