Answer:
The potential difference is  [tex]\Delta V  =  44.40 \  V [/tex]
Explanation:
From the question we are told that
  The diameter of the droplet of oil is [tex]d =  0.71 \mu m  =  0.71 *10^{-6} \ m[/tex]
  The density of the oil is  [tex]\rho =  860 kg/m^3[/tex]
  The distance of separation of the capacitor plate is  [tex]l  = 4.5 \ mm =  0.0045 \  m[/tex]
Generally the radius of the droplet is mathematically represented as
    [tex]r = \frac{d}{2}[/tex]
=> Â Â Â [tex]r = \frac{0.71 *10^{-6} }{2}[/tex]
=> Â Â Â [tex]r = 3.55 *0^{-7} \ Â m[/tex]
Generally the mass of  the oil droplet is mathematically represented as
    [tex]m  =  \rho  *  V[/tex]
Here  V  is the volume of the oil droplet which is mathematically represented as
   [tex]V  =  \frac{4}{3} * 3.142 * (3.55 *0^{-7} )^3[/tex]
    [tex]V  =  1.874 *10^{-19} \  m^3[/tex]
So
    [tex]m  =  860  *  1.874 *10^{-19} [/tex]
=>   [tex]m  =  1.611 *10^{-16} \  kg [/tex]
Generally the electric force acting on the droplet is mathematically represented as Â
    [tex]F  =  E  *  q[/tex]
Here q is the charge on an electron with value  [tex]q =  1.60*10^{-19}\ C[/tex]
This force is equivalent to the weight of the droplet which is mathematically represented as
    [tex]W =  mg[/tex]
So
    [tex]E *  q =  m *  g[/tex]
Here E is the electric field which is mathematically represented as
  [tex]E =  \frac{\Delta V}{l}[/tex]
   [tex]\frac{\Delta V}{l}  *  q =  m *  g[/tex]
=>   [tex]\Delta V  =  \frac{m *  g  *  l }{q}[/tex]
=> Â Â [tex]\Delta V Â = Â \frac{1.611 *10^{-16} Â * Â 9.8 Â * Â 0.0045 }{1.60*10^{-19}}[/tex]
=> Â Â [tex]\Delta V Â = Â 44.40 \ Â V [/tex]
