Question

Suppose that you rub two pennies with a piece of cloth such that they acquire the same charge. How much charge (in C) has to be on each of the pennies such that if they were placed on top of each other, one of them would float 1.0 mm above the other

Answer 1

Complete Question

Suppose that you rub two pennies with a piece of cloth such that they acquire the same charge. How much charge (in C) has to be on each of the pennies such that if they were placed on top of each other, one of them would float 1.0 mm above the other? (hint: think about the forces acting on the hovering penny, and take its mass to be 2.5 grams.)

Answer:

The charge is  $q = 1.65 *10^{-9}\ C$

Explanation:

From the question we are told that

    The required distance between the pennies is $r = 1.0 \ mm = 1.0 *10^{-3} \ m$

    The mass of each penny is  $m = 2.5 \ g = \frac{2.5}{1000} = 2.5 *10^{-3} \ kg$

Generally one penny to float on top of the other , the electrostatic force will be equal to the gravitational force , this is mathematically represented as

       $F_E = F_G$

Here  $F_E = \frac{K * q_1 * q_2 }{r^2 }$

While  $F_g = m * g$

So  

      $\frac{K * q_1 * q_2 }{r^2 } = m g$

Give that the pennies have the same charge we have that $q_1 = q_2 = q$  , so

      $\frac{K * q_1 * q_2 }{r^2 } = m g$

     $\frac{K * q^2 }{r^2 } = m g$

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

        $\frac{9*10^{9} * q^2 }{(1.0 *10^{-3})^2 } = m g$

=>     $\frac{9*10^{9} * q^2 }{(1.0 *10^{-3})^2 } = 2.5 *10^{-3} * 9.8$

=>    $9*10^{9} * q^2 = 2.45*10^{-8}$

=>   $q^2 = 2.722*10^{-18}$

=>   $q = 1.65 *10^{-9}\ C$