Millikan's oil drop experiment direction

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Homework Help Overview

The discussion revolves around Millikan's oil drop experiment, specifically focusing on the forces acting on a charged oil drop when it reaches a state of balance. Participants are exploring the relationship between electric force, weight, and viscous forces in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand the direction of the viscous force acting on the oil drop when it is balanced. Questions are raised about the conditions under which viscous force is present and its relationship to the motion of the drop.

Discussion Status

There is an ongoing exploration of the nature of viscous force in relation to the oil drop's motion. Some participants have provided insights into the behavior of viscous force, while others are seeking clarification on its direction when terminal velocity is achieved. No consensus has been reached yet.

Contextual Notes

Participants are discussing the implications of the oil drop's motion and the forces acting on it, including assumptions about the conditions of motion and the definitions of forces involved.

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Homework Statement



Revered members,
Kindly see my attachment.

Homework Equations





The Attempt at a Solution


When the charged oil drop is balanced, electric force = weight of oil drop.
Eq = mg
Eq = F + U, where F is the viscous force and U is the upthrust force. Since the oil drop attains terminal velocity mg = F + U.
I know viscous force always act upwards. But i can't figure out the direction of viscous force when oil drop is balanced. Please help, members
 

Attachments

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    millikans oil drop.png
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If there is no relative movement (between a body and the medium) there will be zero viscous friction force.
 
Viscous force is directly proportional to velocity.
Viscous force only arises when there is movement of the oil drop (eg. when the electric field is removed and the drop falls. Here viscous force will continue to increase until upthrust + viscous force = weight. Then, resultant force being zero, the drop will fall with terminal velocity)
Also, viscous force always opposes direction of motion.
 
Last edited:
Thanks a lot NascentOxygen and Physics S16 for your replies.
What about the direction of viscous force when terminal velocity is achieved? Will it be upwards?
I think upwards from the statement of Physics S16 that viscous force always opposes direction of motion
 
Yes, always exactly opposite to the direction of [relative] motion.
 

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