This is easy, yet awesome. There are 8 faces to the arrangement of large-metallic plates. Write the values of the charges on each of the faces, left to right.
The charges should be comma separated.
P.S:
1)Use the judge for checking your answer.
2)What is the of Sin(nx)*Sin(mx) from 0-2pi. The answer is a function of m and n.
2. 1/2 [ sin (2pin-2pim) /(n-m) - sin(2pin+2pim)/(n+m) )
ReplyDeletethis is lulz...
ReplyDeletebhaiya,
ReplyDelete1. I am getting 3Q,-2Q,2Q,0,0,2Q,-2Q,3Q
but, it says it wrong.
bhaiya, I used the fact that either we come from left OR right, potential of 0V plate would remain same.
so, V(extreme left)-V(0V)=(x/E)*d
V(extreme right)-V(adjacent)=(x/E)*d
so, the charges in between would be zero as now, p.d. should be zero.
Also, I assumed distance b/w plates as d.
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletetake the sheets to be finite. take the seperation between them as much much smaller than the area.
ReplyDeletethe same is applicable
gud question bhaiya..
ReplyDeleteand about the second one.., isn't it too tough??
@aakash: uske bina hi ho gya.. n i really cudn't get what u wer saying.! :P
@shivanker:
ReplyDeletei meant capacitors ki tarah assume karle, concept of potential define ho jayega, because the body becomes finite
Someone please provide a solution
ReplyDeletehey nice question bhaiya, and thanks to shivanker for explaining me
ReplyDeleteso, irrespective of which plate is earthed, outermost charges are 0?
yes
ReplyDeleteSOMEONE PLEASE GIVE THE SOLUTION
ReplyDeletehi Anonymous.
ReplyDeleteAs Sambhav said, if you earth any plate in the setup, the charges on the outermost plates must be 0.
If it is not so, the outermost plates are at infinite potential. But there is a plate at zero potential close to them. This is not possible.
1.Why does the presence of charge on outermost plates(i think you meant outermost surface the plate is isolated) make their potential infinity
ReplyDeletesee, even the infinity is at 0 potential, and even the second plate is at 0 potential, so, for the first plate, lets calculate potential from both ends
ReplyDeleteE1(infinity)=E2(d), where d is the distance between the two plates
obviouslt RHS is finite, so, for LHS to bi finite, E1 must be small, very small, almost 0
second wali plate ko earth karte hi pure system ka total charge zero ho jayega
ReplyDeleteisliye secong plate pe -4Q carge ayega
AUR ISE AISE BHI DEKH SAKTE HO KI SYSTEM KI ENERGY OUTERMOST PLATE PE JO CHARGE USPE DEPEND KAR RAHI HAI. AGAR OUTERMOST SHELL PE CHARGE HUA TO FIELD BHI HOGI AND ENERGY =(K) FIELD X VOLUME(INFINITE) SO ENERGY WILL BE INFINITE AND AS WE ALL KNOW THAT EVERY SYSTEM TRY TO MINIMISE ITS ENERGY SO OUTER PLATE PE CHARGE 0 HOGA . SINCE CHARGE ON OUTER PLATE = TOTAL CHARGE/2 . THEREFORE TOTAL CHERGE WILL BE ZERO => CHARGE ON SECOND PLATE IS -4Q
Thanks sambhav for the reply. But how can you set the potential at infinity zero when the plates are infinitely large(i hope thats what large meant)and i suppose E1 and E2 are the same in your last comment.
ReplyDeleteAnd btw if you see the class copy you find find the case of two concentric shells where in one case the outer shell was earthed(with the charge on the inner shell as Q) and understandbly the charge on the outer shell was 0. But in the other case when the inner shell was earthed the charge on the outer plate comes out to be Q(1-r/R). Can you draw a parallel between the two situations(in this problem and the shells one.
shells produce rapidly decreasing fields, plates produce constant fields.
ReplyDeleteso, for a shell, there can be charge on the surface. if we assume infinity to be at 0 potential, the potential of the shell is a finite value.