Millikan's experiment with a twist

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The discussion focuses on a modified version of Millikan's oil drop experiment, where the electric field is applied horizontally, causing charged droplets to fall at an angle. The relationship between the forces acting on the droplet is explored, leading to the equation sin(theta) = qE/(b*v), where v represents terminal velocity and b is the drag force upward. Participants clarify that gravity also acts on the droplet, contributing to its downward motion. The upward drag force is defined as bv, with b calculated using the oil drop's radius and the viscosity of air. Overall, the conversation emphasizes understanding the forces and deriving the necessary equations for this variant of the experiment.
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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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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