Answer :
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]
=> [tex]r = \frac{0.71 *10^{-6} }{2}[/tex]
=> [tex]r = 3.55 *0^{-7} \ 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
[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]
=> [tex]\Delta V = \frac{1.611 *10^{-16} * 9.8 * 0.0045 }{1.60*10^{-19}}[/tex]
=> [tex]\Delta V = 44.40 \ V [/tex]