Forces acting on charged oil drops

In summary, the conversation discusses a question about a small sphere moving through air and experiencing a resistive force of kv, given a constant k. The question asks to show that the speed v at which the oil drop moves upwards is determined by the potential difference between the plates, V2 - V1, and is given by v=q/kd x (V2-V1). The conversation also explores the significance of V1, which is determined by the electrical force needed to balance gravity, and how it affects the speed of the oil drop.
  • #1
burno_06
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Hi guys. I'm a new member, just joined today. I've been looking for the answer to the following question for hours, but am unable to do so. Please try to help ASAP! Thanx a million. The question is:

When a small sphere moves through the air with a low speed v it experiences a resitive force given by kv where k is a constant. If the oil drop carries a charge of magnitude q, show that when the potential difference between the plates is V2 the speed v with which the drop moves upwards is given by

v=q/kd x (V2-V1)
 
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  • #2
Welcome to PF!

The rule here is that to get help you must show your work. :smile:

But here's a hint: For the sphere to move at constant speed, what must be the net force on it?
 
  • #3
burno_06 said:
Hi guys. I'm a new member, just joined today. I've been looking for the answer to the following question for hours, but am unable to do so. Please try to help ASAP! Thanx a million. The question is:

When a small sphere moves through the air with a low speed v it experiences a resitive force given by kv where k is a constant. If the oil drop carries a charge of magnitude q, show that when the potential difference between the plates is V2 the speed v with which the drop moves upwards is given by

v=q/kd x (V2-V1)
What is V1? (are we to assume it is the potential for which v = 0?)

AM
 
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  • #4
burno_06 said:
...show that when the potential difference between the plates is V2 ...
I assume that you meant to write the potential difference as V2 - V1, not V2.
 
  • #5
Doc Al said:
I assume that you meant to write the potential difference as V2 - V1, not V2.
I think V1 has to be the voltage for which v = 0. Otherwise you would have to know the mass of the oil drop.

AM
 
  • #6
Andrew Mason said:
I think V1 has to be the voltage for which v = 0. Otherwise you would have to know the mass of the oil drop.
Not sure what you mean. What determines the terminal velocity is the strength of the electric field, which depends on the potential difference. (Assuming that we can ignore gravity, the mass is not needed.)
 
  • #7
Doc Al said:
Not sure what you mean. What determines the terminal velocity is the strength of the electric field, which depends on the potential difference. (Assuming that we can ignore gravity, the mass is not needed.)
I don't think we can ignore gravity. In order to keep the oil drop still, you will need to apply an electrical force to the oil drop to balance gravity, so some potential difference to the plates is needed. That is V1. The speed will be determined by the electric force (qE=qV/d) in excess of that.

AM
 
  • #8
OK. I'll buy that. :smile:
 

FAQ: Forces acting on charged oil drops

1. What are the forces acting on charged oil drops?

The forces acting on charged oil drops are electrical forces, gravitational forces, and air resistance.

2. How do electrical forces affect charged oil drops?

Electrical forces are responsible for the repulsion or attraction between charged oil drops. Like charges repel each other, while opposite charges attract.

3. What role does gravity play in the motion of charged oil drops?

Gravity is responsible for the downward force on charged oil drops, which causes them to accelerate towards the ground. This force is typically much smaller than electrical forces in the case of small oil drops.

4. Why is air resistance important in the study of charged oil drops?

Air resistance or drag is the force exerted by air molecules on a moving object. It affects the motion of charged oil drops and must be taken into account when studying their behavior in an experiment.

5. How can the forces on a charged oil drop be balanced?

The forces acting on a charged oil drop can be balanced by adjusting the charge or mass of the drop. By carefully controlling these variables, it is possible to create a state of equilibrium where the oil drop remains suspended in mid-air.

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