A charged oil drop of radius R and density D is held stationary in a vertically downward electric field of 1.65*10^6.find the magnitude and sign of the charge???
I am concerned about how do you find the sign of the drop and not the magnitude coz i aldready found it ..pls help???

Question

A charged oil drop of radius R and density D is held stationary in a vertically downward electric field of 1.65*10^6.find the magnitude and sign of the charge???  
I am concerned about how do you find the sign of the drop and not the magnitude coz i aldready found it ..pls help???

mani_jha · Answer

The oil drop will be held stationary only when the coulomb force on it balances its weight. Find the weight(volume*density) and the force is mass*field.
Equate both and find the charge

anonymous · Answer

i know how to find the charge
"I am more concerned with the sign???

anonymous · Answer

for equilibrium ,
the net force on the oil must be equal to zero.Here the gravitational force in downward direction hence electric force must be in upward direction.but electric field is in downward direction therefore the charge must be negative...and magnitude is given by following equation..
                     i.e Fe=Fg=>Q*E=M*g=>Q=(((4/3pi*R^3)*d)*g)/E.put the value in this equation and get the magnitude of charge...

anonymous · Answer

so the charge should be negative???

anonymous · Answer

@sarkar,if the charge is positive then electric force and electric field intensity vector will be in same direction and if it is negative then these will be in opposite direction.

anonymous · Answer

yes ,charge should be negative....

anonymous · Answer

okay @taufique thanks for explaining me crystal clearly .

anonymous · Answer

welcome bro....

anonymous · Answer

@taufique check my question on S.H.M.

anonymous · Answer

say me..

anonymous · Answer

i have posted my question a little bit earlier??

anonymous · Answer

http://openstudy.com/study#/updates/4f634be7e4b079c5c631b05e

anonymous · Answer

ok..