A charged oil drop is suspended in uniform field of 3 x 10^4 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/s^2) (a) 3.3 x 10^-18 C (b) 3.2 x 10^-18 C (c) 1.6 x 10^-18 C (d) 4.8 x 10^-18 C

Given

Electric field, E = 3 x 104 V/m

Mass of the drop, M = 9.9 x 10-15 kg

Let us consider q as the amount of charge that the drop carries

The coulomb force balances the gravitational force acting on the drop at equilibrium

Hence,

qE = Mg

q = Mg/E

q = (9.9 x 10-15 x 10)/3 x 104

We get,

q = 3.3 x 10-18

Therefore, the correct option is (a)

Was this answer helpful?

 
   

0 (0)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question