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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