Question

In the Millikan’s oil drop experiment the oil drop is subjected to a horizontal electric field of 4N/C and the drop moves with a constant velocity making an angle ${45}^0$ to the horizontal. If the weight of the drop is W. The charge on the drop is (neglect buoyancy)

• A. W
• B. W/4
• C. W/2
• D. 3W/4

Answer 1

The correct option is B W/4

As the resultant makes angle ${45}^{\circ }$ with the Horizontal,
$V_1=V_2=V$
In Horizontal direction,
$4q=\left(6\pi \eta r\right)V$ ------------------ (I)
Similarly in vertical direction,
$W=\left(6\pi \eta r\right)V$ ------------------ (II)
From (I) and (II), we easily see that,
$4q=W$
$\Rightarrow q=\frac{W}{4}$

So, the answer is option (B).