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

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SUMMARY

The discussion focuses on calculating the viscous force acting on an oil drop in Millikan's experiment when subjected to different electric fields. In the first scenario, the oil drop experiences a downward force due to gravity and an upward electric force, leading to the equation qE + F(viscous force) = mg. In the second scenario, with a horizontal electric field, the drop moves at a 45-degree angle, resulting in the conclusion that the viscous force is mg/2. The key takeaway is that the viscous force can be expressed using the formula F = 6πrηv, where η represents the fluid's viscosity and v the velocity of the drop.

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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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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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