electrically charged drops of mercury fall from altitude h into a spherical metal vessel of radius R. in the upper part of which there is a small opening. The mass of each drop is m & charge is Q.What is the number 'n' of last drop that can enter the sphere.Given that (n+1)th
drop just fails to enter the sphere?
49balance forces! will try to come up with a solution in few hours!!!
btw akhil..u in 11 or 12???
1The nth particle enters just the container because the weight of the particle is more than the total electrostatic repulsive force ...
I'll try to get a simplifed mathematical expr. after sometime ...
49let the electrc field at the hole due to n drops be E
thus balancing force on the (n+1)th drop we get,
QE=mg
rest of the problem is very easy .. only we need not to get confused!!
total charge in n drops = nQ
so since the shell is metallic, and the hole is small such that no electric lines of force escapes [i am trying to cheat in order to work less!! ;)]
so charge induced in inner surface is (-nQ)
so charge appearing on outer surface is nQ
field on the hole is nQ8πεoR2 [not nQ4πεoR2...why??]
thus equating we get...
mg=nQ28πεoR2
that gives,
n=8πεoR2mgQ2
49why is the field at the hole nQ8πεoR2 and not nQ4πεoR2 ???
lets see who can answer this? ;)
49now lets try out some googlies here!!
WHAT IF THE SPHERE IS NON-CONDUCTING??
take each drop is spherical and has a radius of r ... !!
hint: i can think about nothing else other than using solid angle concept!
66@subho
balancing forces won't give you the desired result. Can you figure out why?
1Anant sir .... just a doubt ..... the electric field intensity due to one falling charge on another inside the container is not const . but changing .... as the distance between the 2 is continuosly decreasing ... [ charges r not static ]
So how can we apply the normal algebraic eqn. ????
