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?

Respuesta :

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

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

   

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

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

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

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

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]