Question

two 200 pound lead balls are separated by a distance 1m. both balls have the same positive charge q. what charge will produce an electrostatic force.between the balls that is of the same order of magnitude as the weight of one ball?

Answer 1

Answer:

The  charge is  $q = 3.14 *10^{-4} \ C$

Explanation:

From the question we are told that

     The mass of each ball is  $m = 200 \ lb = \frac{200}{2.205} = 90.70 \ kg$

       The distance of separation is  $d = 1 \ m$

Generally the weight of the each ball is mathematically represented as  

      $W = m * g$

where g is the acceleration due to gravity with a value $g = 9.8 m/s^2$

substituting values

      $W = 90.70 * 9.8$

      $W = 889 \ N$

Generally  the electrostatic force between this balls is mathematically represented as

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

given that the the charges are equal we have

    $q_1= q_2 = q$

So

         $F_e = \frac{k * q^2 }{d^2}$

Now from the question we are told to find the charge when the weight of one  ball is equal to the electrostatic force

So  we have

       $889 = \frac{9*10^9 * q^2}{1^2}$

   =>   $q = 3.14 *10^{-4} \ C$