| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Jun 2007 18:16:39 IST
|
|
|
the questions r as follows:
|
|
|
|
16 Jun 2007 18:33:57 IST
|
|
|
there will not b any electric field after r=R.
u can prove this using gauss law
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
16 Jun 2007 18:41:46 IST
|
|
|
can u prove it.??
|
SOMETIMES I WISH I COULD TURN BACK TIME. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
16 Jun 2007 18:44:17 IST
|
|
|
q.2 three concentic conducting spherical shells of radius r1 r2 r3 and charges q1 q2 q3 resp. innermost and outermost shells are earthed. then,
a q3/q2 = -1/3
b q1 = -q2/4
c q3/q1 = 3
d q1+ q3 = -q2
|
SOMETIMES I WISH I COULD TURN BACK TIME. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
16 Jun 2007 18:47:24 IST
|
|
|
the graph ques
after r=R using gauss law the net charge enclosed is zero. hence no elec field
|
this reply: 2 points
(with 0
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
16 Jun 2007 18:54:36 IST
|
|
|
as inner and outer are earthed they r at same potential.
hence potential diff between them is 0.
kq3/r3+kq2/r3+kq1/r3=kq3/r3+kq2/r2+kq1/r1
|
this reply: 2 points
(with 0
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
17 Jun 2007 17:17:33 IST
|
|
|
when inner most and outermost shells are earthed...there POTENTIAL becomes 0...
now equate their potentials to 0 and get the answers
note:charge on the innermost and outer most spheres are changed but the charge on middle shell remains same....
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
17 Jun 2007 17:21:27 IST
|
|
|
For the 1st Q...
consider a gaussian surface through the small thickness of the shell...since electric field lines can't pass through a conductor...E=0.
using gauss law...enclosed charge =0....so a charge -q is induced at the inside surface of shell...using conservation of charge on shell...charge on outermost surface is 0...
so there doesn't exist field outside R.
note: charge always resides on the surface of a conductor...
|
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}
![[Thumb - untitled.JPG]](/upload/2007/6/16/f3e5887229b71261e61976ac6b416245_20455.JPG_thumb)
