What is the viscous force acting on an oil drop in Millikan's experiment?

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Homework Help Overview

The discussion revolves around the viscous force acting on an oil drop in Millikan's oil drop experiment, particularly when subjected to different electric fields. The original poster describes a scenario where an oil drop moves downward under a vertical electric field and then at a 45-degree angle when a horizontal electric field is applied.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the oil drop in both scenarios, questioning the assumptions about the viscous force and its consistency across the two cases. There are discussions about the direction of motion and the implications of the 45-degree angle on the forces involved.

Discussion Status

Some participants have offered insights into the forces acting on the drop, while others have raised questions about the assumptions made regarding the viscous force and its potential variation between the two cases. The conversation reflects a productive exploration of the problem without reaching a definitive conclusion.

Contextual Notes

Participants note the neglect of buoyant force due to air and the implications of the electric fields on the motion of the drop. There is an acknowledgment of the need to clarify the relationship between the velocities in both scenarios to further understand the viscous force.

Krushnaraj Pandya
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Homework Statement


In Millikan's oil drop experiment on applying a vertically upward electric field an oil drop (of mass m) moves vertically downward with certain terminal speed. On applying double the electric field in horizontal direction, the drop moves making 45 degrees with the vertical. Neglecting buoyant force due to air, what is the viscous force acting on the drop in first case?

Homework Equations


Fluid mechanics+electrostatics+mechanics equations

The Attempt at a Solution


Attached below. From first case, qE+F(viscous force)=mg, from second case F/root2 =mg but this is inconsistent and the answer is mg/2. I'd be grateful for some help
firstcase.jpg
case2.png
 

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  • case2.png
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Without the horizontal field it was moving downwards. In your diagram, after adding a horizontal field you have it moving upwards. Does that seem reasonable?
 
haruspex said:
Without the horizontal field it was moving downwards. In your diagram, after adding a horizontal field you have it moving upwards. Does that seem reasonable?
The arrow shows the viscous force, I have it moving downwards. Note the separate arrow pointing southeast shows the velocity.
 
Krushnaraj Pandya said:
The arrow shows the viscous force, I have it moving downwards. Note the separate arrow pointing southeast shows the velocity.
Sorry, I should have read it more carefully.
Are you assuming the viscous force is the same in both cases?

In the second diagram, isn't the vertical electric force still acting?
However, to get the answer mg/2 I have to assume it is no longer acting.
What does the fact that it moves at 45 degrees tell you about the applied forces in the second case?
 
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haruspex said:
Sorry, I should have read it more carefully.
Are you assuming the viscous force is the same in both cases?
the viscous force is 6pi*r*n*v. I noticed just now that the velocity might be different in both cases. How can we obtain a relation between them?
 
Krushnaraj Pandya said:
the viscous force is 6pi*r*n*v. I noticed just now that the velocity might be different in both cases. How can we obtain a relation between them?
See my later edit above.
 
Oh, right- I got the correct answer by equating forces in vertical and horizontal directions and plugging into the first. My main mistake was that I considered F to be the same in both cases, Thank you very much for your help :D
 

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