Question

Two positive point charges, each with charge q, separated by a distance d, repel each other with a force of magnitude 20 N. What is the magnitude of the force between two positive point charges of magnitude 11.11 q, separated by a distance 2.5 d in units of N

Answer 1

Answer:

The value is  $F_2 = 395 \ N$

Explanation:

From the question we are told that

   The magnitude of the charge of each positive charge for the first case is $q_1 = q_2 = q$

   The distance between the charges for the first case is  $d$

   The  force between the charges for the first case  is  $F = 20 \ N$

     The magnitude of the charge of each positive charge for the second case is $q_1 = q_2=11.11q$

     The  distance between the charge for the second case is $2.5d$

 

Generally for the first case the force between the charge is mathematically  represented as

       $F_1 = \frac{k * q^2 }{d^2}$

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

  So  

     $20= \frac{k * q^2 }{d^2}$

Generally for the second  case the force between the charge is mathematically  represented as

       $F_2 = \frac{k * (11.11q)^2}{(2.5d)^2}$

       $F_2 = \frac{k * 11.11^2 *q^2}{2.5^2d^2}$

=>      $F_2 = \frac{11.11^2}{2.5^2} F_1$

=>      $F_2 = \frac{11.11^2}{2.5^2} * 20$

=>      $F_2 = 395 \ N$