# Forces acting on charged oil drops

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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Doc Al
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Welcome to PF!

But here's a hint: For the sphere to move at constant speed, what must be the net force on it?

Andrew Mason
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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

Last edited:
Doc Al
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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.

Andrew Mason
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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

Doc Al
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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.)

Andrew Mason
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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

Doc Al
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