Millikan's experiment with a twist

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The discussion centers on a modified version of Millikan's oil drop experiment, where the electric field is applied horizontally instead of vertically. This configuration results in the charged droplets falling at an angle theta with respect to the vertical. The relationship derived is sin(theta) = qE/(b*v), where v represents the terminal velocity, b is defined as 6*pi*n*a (with a being the droplet radius and n the air's viscosity coefficient), and qE is the horizontal force acting on the droplet.

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The Millikan's oil drop experiment is doen with the electric field horizontally, rather than vertically, giving the charged droplets an acceleration in the horizontal direction. The result is that the droplet falls in a straight line which makes an angle theta with the vertical. Show that sin(theta)= qE/(b*v), where v is the terminal velocity. (bv is the drag force upward).

What are the forces acting on the oil? Just the horizontal force, right? So qE=ma, how do I get it to look like sin(theta)= qE/(b*v)?

Thanks.
 
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What is b? Also wouldn't gravity also be acting on the oil drop since it's falling?
 
bv is the upward force, and b=6*pi*n*a, where a is the radius of the oil drop and n is the coefficient of viscosity of the air.
So there is also a velocity downward?
 

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