Direction of viscous force in Millikan's oil drop experiment

Thread Starter

logearav

Joined Aug 19, 2011
243
Revered Members,
In millikan's oil drop experiment, the motion of oil drop is due to gravitational force and then the motion is due to electric field.
In the first case, when the oil drop moves downwards due to gravitational force, the viscous force acts in upward direction opposing its motion.
Now, electric field is switched on, and if the upper plate is +ve and the lower plate is -ve, the oil drop will move upward, assuming the charge of oil drop to be negative. Now what about the direction of viscous force? I feel, viscous force's direction is downward. Am I right?
Is it true, the direction of viscous force is always opposite to the direction of motion of substance?
 

Thread Starter

logearav

Joined Aug 19, 2011
243
Ok russ_hensel. Now I want to know what is the direction of viscous force when the oil drop is balanced. I think when the forces are balanced the oil drop won't move. So viscous force will be zero, since viscous force is directly proportional to velocity. Am I inferring right?
 

thatoneguy

Joined Feb 19, 2009
6,359
I'm not sure if it's a translation error or incorrect terms.

The Oil Drop Experiment was to find an electric field force equal but opposite to the force of gravity.

The only place viscosity comes into play is to verify the oil drop is an actual sphere, which a viscous fluid in air with no net forces will assume.
 

Thread Starter

logearav

Joined Aug 19, 2011
243
Revered thatoneguy,
Please see my attachment. According to Millikan Oil drop experiment, a charged oil drop means that the oil drop moves in an electric field. X rays are passed to ionise the air surrounding the oil drop and some electrons got attached to oil drop making it negative and the oil drop moves upward if the electric field is applied(Assuming the upper plate is positive and the lower plate is negative, due to applied electric field). If the forces are balanced,i.e Electric force and the drag force, i suppose the oil drop won't move. In that case viscous force would be zero, according to my reasoning. But one of my teacher says the viscous force would always act opposite to the motion of oil drop and since the oil drop moves upward, the viscous force would act downward.
He reasons that balanced oil drop means the net forces acting on the drop is zero and the oil drop still moves but with constant terminal velocity so viscous force will still come into play.
I solicit the help from revered members.
 

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