bhaiya, hum electric field jo consider kar rahey hai woh to dusri surface ki hai na, so we should take it as sigma/2E.
@shivanker, can you tell me the approach of potential difference-cant get it. If you are referring to as what is done in motional emf then, actually there also we are balancing the forces.
yaar i ws saying ki we basically apply the condition for equilibrium of charges *inside* the sheet..
as for those on the surface.. consider just *one* of them(i mean just one discrete charge) instead of the whole surface density.. and u'll get the same answer.. ie, B.v.e(o)
hey shivanker, though I got that eq is attained inside the conductor but, what you have said that consider force on one discrete charge, I think supports sigma/2E because while considering that force, we wont be adding up the field due to that discrete charge which is also sigma/2E. So, does that mean ki us charge ki tendency wapas plate ke andar jaane ki hai. Please somebody explain!!
you said na that we can consider equilibrium wrt a discrete charge on the surface. if we are considering that then we would be taking electric field by rest of the sheet which is sigma/2(epsilon) and not sigma/epsilon.
jo hum electric field sigma/epsilon kehte hain woh pure conductor ki hoti hai. one sheet is not a conductor, one plate(having 2 surfaces) is a conductor. Also take the example of disc where field is sigma/2epsilon if x tends to zero
leave it, balancing force between the plates is right.that's fine.actually, you might be considering that charge is small but, distance where field is considered is also very small so, actually it is somewhat indeterminant i.e. contribution cant be neglected.
is the answer B*V*(epsilon)
ReplyDeletesame as aakash..
ReplyDeleteyup same!
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeleteI am getting twice of above all.
ReplyDeletehey somebody explain this to me please!
ReplyDeleteI am getting 2*B*V*epsilon
ReplyDeletemethod is in equilibrium, net force acting on any surface=0.
therefore, (sigma/2E)*(sigma*A)=(sigma*A)*V*B
yes,I also did same as amit and bhaiya edited my post so, may be it is correct
ReplyDeleteoh, thanks amit
ReplyDeleteforce on charge due to it's own field....strange...
ReplyDeleteyaar all electrons dont come to the surface. toh, we take the force on them due to the sheet
ReplyDeletebhaiya plz confirm d answer..
ReplyDeleteequating forces.. we get 2Bv(e0)
equating potential differences.. we get Bv(e0)
I guess field at the surface of a conductor is sigma/epsilon0.
ReplyDeletemaybe its not the right place, but thanks for the integration question of e^(-x^2). aaj ts mein matrix match type mein aa gaya tha.
ReplyDeletebhaiya, hum electric field jo consider kar rahey hai woh to dusri surface ki hai na, so we should take it as sigma/2E.
ReplyDelete@shivanker, can you tell me the approach of potential difference-cant get it. If you are referring to as what is done in motional emf then, actually there also we are balancing the forces.
no amit.. i ws just equating the p.d.'s
ReplyDeleteie, |E field|.d = B.v.d
where d is the thickness of the plate..
i think that's wrong...according to me, field 'inside' the conductor should be 0...
ReplyDeleteu mean.. the forces??
ReplyDeletecz field to ek hi h.. zero kaise ho sakti h..?
anyways.. got it bhaiya.. answer vse bhi same hi aa rha h.. :P
ReplyDelete@amit: we need to balance the force on the charges in the bulk of the plate.. and not the ones on the surface!
shivanker, couldn't get you. Finally what are you getting? And also if equilibrium is attained, it is for every charge.
ReplyDeleteyaar i ws saying ki we basically apply the condition for equilibrium of charges *inside* the sheet..
ReplyDeleteas for those on the surface.. consider just *one* of them(i mean just one discrete charge) instead of the whole surface density.. and u'll get the same answer.. ie, B.v.e(o)
hey shivanker, though I got that eq is attained inside the conductor but, what you have said that consider force on one discrete charge, I think supports sigma/2E because while considering that force, we wont be adding up the field due to that discrete charge which is also sigma/2E. So, does that mean ki us charge ki tendency wapas plate ke andar jaane ki hai. Please somebody explain!!
ReplyDeleteedit:the tendency is to come out of the plate which is not possible so, may be that is the reason.But, please confirm.
ReplyDeletedidnt get u amit..
ReplyDelete"we wont be adding up the field due to that discrete charge which is also sigma/2E"
field due to that single discrete charge is sigma/2e ?????
you said na that we can consider equilibrium wrt a discrete charge on the surface. if we are considering that then we would be taking electric field by rest of the sheet which is sigma/2(epsilon) and not sigma/epsilon.
ReplyDeleteso, answer would come twice of actual answer according to this approach. Tell if I am going wrong somewhere!
ReplyDeletearey field would be due to the charges on the same surface as well as the field due to the opposite surface.. won't it?
ReplyDeletejo hum electric field sigma/epsilon kehte hain woh pure conductor ki hoti hai. one sheet is not a conductor, one plate(having 2 surfaces) is a conductor. Also take the example of disc where field is sigma/2epsilon if x tends to zero
ReplyDeletearey to yaar "poore conductor" mei se ek discrete charge, one electron, nikalne se kya kaafi change aa jayega??
ReplyDeletewe are taking that discrete charge from surface so, yes it would-i think similar as an infinite small cavity
ReplyDeletesorry amit.. m unable to get u..
ReplyDeletend even m not sure about what m saying..
leave it, balancing force between the plates is right.that's fine.actually, you might be considering that charge is small but, distance where field is considered is also very small so, actually it is somewhat indeterminant i.e. contribution cant be neglected.
ReplyDelete