Question

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.
Required:
What is the charge on the dust particle?

Answer 1

Answer:

The  charge on the dust particle is  $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

    The length is  $l = 2.0 \ m$

    The width is  $w = 4.0 \ m$

   The charge is  $q = -10\mu C= -10*10^{-6} \ C$

    The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

   

Generally the electric field on the carpet is mathematically represented as

           $E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

           $E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

           $E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        $F__{E}} = F__{G}}$

=>     $q_d * E = m * g$

=>      $q_d = \frac{m * g}{E}$

=>      $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=>     $q_d = 6.94 *10^{-13} \ C$