33[4] Full test II
14 was easy but how to do questions 15 and 16?,
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14 Answers
33wellll.......
16
|-q q | Q-q -Q+q |
| | |
| | |
Now V of left=V of right =0
V of middle = qd/Aε0 (from left)
V of middle calculationg from right... =(Q-q)3d/Aε0
both are same...
solve this.. u will get...q=3Q/4
therefore charge on right plate=-Q+q =-Q/4
initially it was zero.. so Q/4 will go to earth
(the charge on outer surface of plate is zero for grounded plate... Sir once told this.. don't know how.. if any one can tell about this...[7])
33for 15th.. also charge on outer part of left plate will become zero... inner side remains same here... as no change in left part of middle plate..
so all Q/2 charge will go to earth...
1i have a doubt in a similar question,there are three plates with equal spacing btwn them
like this
[+Q] [+7Q] [+10Q]
now the middle one is grounded,what will be the charge distrib on the left plate?
62x||Q-x x1||x2 y||10Q-y
now try that the electric filed is zero between the plates...
and then also try to put potential between the middle plate as zero.
62ok let me try
x=Q-x+x1+x2+10Q
2x=11Q+x1+x2
Q+x1=x2+10Q
x1=x2+9Q
Q+x1+x2+y=10Q-y
x1+x2+2y=9Q
now work on a charge on bringing it from infinity should be zero..
so charge on both sides from infinity should be zero!!!!
so Q+x1+x2+10Q=0!
There can be one more equation but I am assuming that it will be redundant!
62It was a very good question vivek..
I guess even I had never solved one problem of this type before!