-->

Mechanical Properties Of Fluids

Question
CBSEENPH11026183

A charged oil drop is suspended in uniform field of 3 x 104 V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9 x 10-15 kg and g = 10 m/s2 )

  • 3.3 × 10-18 C

  • 3.2 × 10-18 C

  • 1.6 × 10-18 C

  • 4.8 × 10-18 C

Solution

A.

3.3 × 10-18 C

In steady state, 

electric force on drop = weight of drop 

 q E =  mg

  q = mgE          = 9.9 × 10-15 × 103 × 104  q  = 3.3 ×10-18 C