A 0.71um-diameter droplet of oil, having a charge of +e, is suspended in midair
ID: 1687846 • Letter: A
Question
A 0.71um-diameter droplet of oil, having a charge of +e, is suspended in midair between two horizontal plates of a parallel-plate capacitor. The upward electric force on the droplet is exactly balanced by the downward force of gravity. The oil has a density of 860 (kg/m^3), and the capacitor plates are 6.0 mm apart. What must the potential difference between the plates be to hold the droplet in equilibrium? What are the units for solving deltaV ?A 0.71um-diameter droplet of oil, having a charge of +e, is suspended in midair between two horizontal plates of a parallel-plate capacitor. The upward electric force on the droplet is exactly balanced by the downward force of gravity. The oil has a density of 860 (kg/m^3), and the capacitor plates are 6.0 mm apart. What must the potential difference between the plates be to hold the droplet in equilibrium? What are the units for solving deltaV ?
Explanation / Answer
the force acting on a charge 'e' in an electric field 'E' is given by F = Ee
In parallel plate capacitor the electric field is E =V/d, where 'V' is potential and 'd' is distance between to plates the force due to electric field acting on liqued drop is Fe = Ve/d-----------(1) the force due to gravity on droup is Fg = m g -----------(2) these two forces equal ,that is Fe = Fg Ve/d = m g V = m g d/e = (? 4/3 p R3 g d )/e = (860*4/3*3.14*0.3553 *9.8 *6*10-3) /1.602*10-19 V = 580.02v therefore the potential defference between two plates is 580v.
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