1
dimensions (dimentime)
·2008-12-03 23:35:31
sir, is the system in equilibrium,
if yes.then there has to be a resistance in the circuit, otherwise the charge on capacitance becomes constant, ie, q = CE so, dq/dt becomes zero,
pls reply.
62
Lokesh Verma
·2008-12-03 23:37:31
The system is not in equilibrium. I have asked for the acceleration (so how can the sytem be in equilibrium ;)
1
dimensions (dimentime)
·2008-12-03 23:48:47
sir, my acc. of block is comming constant
a = E2ε0b(k-1)/md
& velocity = {E2ε0b(k-1)/md}t
please verify........................
1
Rohan Ghosh
·2008-12-03 23:54:40
here goes the full solution
writing the constraints for both
we have accln of lower block= accln of upper slab=a
then
mg-2T=T+F
T=mg-f/3
where F force due to capacitor
F=-dU/dx
capacitance=εA/d(x/b+k(b-x)/b)
dC/dx= εA/bd(1-k)
F=-E2dC/dx
=-E2εA(1-k)/2bd
hence acceleration=
((mg-F)/3+F)/m
=mg+2F/3m
current =
EdC/dt= EdC/dx*dx/dt
=EεA(1-k)/bd*v
but v=at
so the answer follows
earlier i made a short mistake
1
dimensions (dimentime)
·2008-12-03 23:58:26
& current is comming,
i = {E5(ε0b(k-1))2/2d2m}t
pls. verify...........
1
Rohan Ghosh
·2008-12-04 00:33:53
and @dimensions
how come u re not getting g in your answers and A = area of the capacitors?
1
dimensions (dimentime)
·2008-12-04 00:49:58
here A will be
A = b*l
now l is getting cancled in my solution & my ans is comming in terms of b.........
9
Celestine preetham
·2008-12-04 04:38:27
as u dint menton mass of block i took it as m
i = εEb(k-1)at/d
a = g - εbE2(k-1)/md
dimensions dint consider g
1
Akand
·2008-12-13 11:05:21
So wats d answer for this??????????