A small charged bead has a mass of 3.6 g. It is held in a uniform electric field E = (200,000 N/C, up). When the bead is released, it accelerates upward with an acceleration of 24 m/s^2. What is the charge on the bead?

Respuesta :

Answer:

The charge on the bead is  [tex]q = 6.084 *10^{-7}\ C[/tex]

Explanation:

From the question we are told that

     The  mass of the bead is  [tex]m = 3.6 \ g = 0.0036 \ kg[/tex]

      The  magnitude of the electric field is  [tex]E = 200,000 \ N/C[/tex]

       The  acceleration of the bead is  [tex]a = 24 m/s^2[/tex]

Generally,  the  electric force on the bead is  mathematically represented as  

         [tex]F_ e = q E[/tex]

Where q is the charge on the bead

   Now the gravitational force opposing the upward movement of the bead is  mathematically represented as

       [tex]F_g = mg[/tex]

Generally the net force on the bead is  mathematically represented as

       [tex]F = F_e - F_g = m* a[/tex]

=>    [tex]qE - mg = ma[/tex]

Now substituting values

       [tex]q * 200000 - 0.0036 *9.8 = 0.0036 * 24[/tex]

     [tex]q = 6.084 *10^{-7}\ C[/tex]