Question

Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb. find the separation between the balls if the lower ball is restrained from moving.​

Answer 1

Answer:

The distance of separation is $d = 0.092 \ m$

Explanation:

The mass of the each ball is  $m= 10 g = 0.01 \ kg$

 The negative charge on each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Now we are told that the lower ball is  restrained from moving this implies that the net force acting on it is  zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

          $F = \frac{kq_1 * q_2}{d}$

=>       $m* g = \frac{kq_1 * q_2}{d}$

here k the the coulomb's  constant with a value  $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

So  

      $0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}$

            $d = 0.092 \ m$