Question

In Millikan's oil drop experiment, if the electricfield between the plates was of just the right magnitude, it wouldexactly balance the weight of the drop. Suppose a tiny sphericaloil droplet of radius 1.6 X 10-4 cm carries a chargeequivalent to one electron. What electric field is required tobalance the weight? (The density of oil is 0.85 g/cm3,e= 1.6 X 10-19 C.)

Answer 1

Answer:

The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb

Explanation:

In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, $|F_{electrostatic}| = |F_{gravitational}|$

where

$F_{electrostatic}=q.E$

where q is the charge of the droplet and E is the electric field. On the other hand

$F_{gravitational}=m.g=V.d.g=\frac{4}{3}.\pi.r^3.d.g$

where m,V ,d and r  are the mass, volume, density and radius of the oil droplet respectively and g is the gravitational acceleration (g=9,80665 m/sn^2). By using the first equation and solving it for the electric field we can write,

$q.E=\frac{4}{3}.\pi.r^3.d.g\\E=\frac{4}{3}.\pi.\frac{r^3.d.g}{q}\\E=\frac{4}{3}.\pi\frac{(1,6x10^{-4})^3.(0,85x10^{-3}).(9,81)}{1,6x10^{-19}}\\E\approx 3,49x10^{5} (N/C)$

(Note: d=0.85 x 10^-3 kg/cm^3 and the unit of electric field is Newton per coulumb (N/C))