| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 May 2007 11:13:27 IST
|
|
|
due to the charge q placed in the shell, redistribution of charges will take place . - q will appear on inner surface of shell and on outer surface +q . Now , if u apply gauss law to find E at a point , now also the charge enclosed will be q . so , there will be electric field . So , assertion is wrong !!!
|
Manasi....
NIT-Allahabad...
............................................................
Challenges are High, Dreams r New..
The World out thr is waiting for U !!
Dare to dream, Dare to Try..
No Goal is distant, no Star is too high !!! |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
17 May 2007 13:44:38 IST
|
|
|
E at a point OUTSIDE the conductor will be zero as the enclosed charge is Q...
but inside the conductor...gaussiun syrface is considered through the conductor...whose enclosed charge is 0...and field b/w the inner and outer surfaces is 0 while there is field inside the cavity...
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
17 May 2007 14:32:34 IST
|
|
|
yup totally agree with neeraj
|
     
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
17 May 2007 21:59:43 IST
|
|
|
in case the point is inside the conductor then thr will be no NET electric field as charge enclosed will be 0... but still defintely electric field lines will exist, but since they will be two opposite charges, they will nullify each other...
|
Manasi....
NIT-Allahabad...
............................................................
Challenges are High, Dreams r New..
The World out thr is waiting for U !!
Dare to dream, Dare to Try..
No Goal is distant, no Star is too high !!! |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
18 May 2007 13:33:52 IST
|
|
|
Since there can't be electric field inside a conductor...THEREFORE..on applying gauss law for a surface passing through the conductor...
Since lhs of gauss law is 0..so rhs should also be 0 ...so a -q is induced on the inner surface...but the charge on the conductor has to be 0...so a +q is induced at the outer surface...
the field lines of the +q which was initially placed emerge radially outwards and END at the innere surface of the shell(at -q)
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
21 May 2007 18:59:32 IST
|
|
|
I asked this que. to my teacher and got a satisfactory reply - E inside the conducting material is zero . It is zero inside the hollow conductor only when the charge enclosed is zero . And simply , just apply gauss laws and u get E = non zero . I had given this answer , but others confused me by saying E inside conductor is zero . I hope u all agree with my solution !!!!!!
|
Umang |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
21 May 2007 19:18:39 IST
|
|
|
umang yaar main kab kah raha hoonki tum theek nahi ho but main kah raha hoon e =0 in body of conductor not in cavity .
|
adversities cause some men to break other to break records............i m of the other type....... :-) |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
21 May 2007 19:22:32 IST
|
|
|
Hey ankur ! yes , u r right !! U just see , it was a very easy question and still it took so long to get an answer !!!!! It proves that we hav a lot more to learn !!!
|
Umang |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
2
