Question

In the Millikan's experiment, the oil drop is subjected to a horizontal electric field of 2N/C and the drop moves with a constant velocity making an angle of ${45}^0$ with the horizontal. If the weight of the drop is W, then the electric charge, in coulomb on the drop is

• A. W
• B. W/2
• C. W/4
• D. W/8

Answer 1

The correct option is B W/2

Let the change on drop be x,
As, the angle Q is ${45}^0$
$v_1=v_2$
As, $v_1=v_2$, the viscous force experienced by the body in both directions should be same, As, the body is moving with constant velocity the force acting should be 0.
In X direction.
$2x=$ viscous force --------(i)
in y directions,
$W=$ viscous force ---------(ii)
Equating (i) and (ii) we get,
$x=\frac{w}{2}$

So, the answer is option (B).