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![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jun 2008 21:01:15 IST
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A hollow sphere of radius 2R is charged to V volts and another smaller sphere of radius R is charged to volt V/2 . Then the smaller is placed inside the bigger spher without changing the net charge on each sphere. the petential differenbce between the 2 spheres would be??
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2 Jun 2008 21:04:27 IST
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is the answer 3/4 V
temme if m correct
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2 Jun 2008 21:06:16 IST
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bro i m also getting the same one but it is v/4...but plzzz tell me ur tactics to solve this q. plzzz
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2 Jun 2008 21:13:16 IST
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ITS 3/4V.........
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DRACO DORMIENS NANQUM TITILANDUS !!!! |
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2 Jun 2008 21:18:33 IST
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even i am gettin 3V/4
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FAILURE, THE FIRST STEP TO SUCCESS |
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2 Jun 2008 21:24:48 IST
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ok now tell me how????
even i m also getting it....plzzz explain thoroughly...plzzz
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2 Jun 2008 21:31:30 IST
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V/4 ?
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2 Jun 2008 21:34:21 IST
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ya tell me the solution plzzzz
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2 Jun 2008 21:38:23 IST
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how did u get 3v/4?
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2 Jun 2008 21:46:51 IST
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i have answered here
http://iit-redefined.zzl.org/viewtopic.php?f=7&t=481&p=2716#p2716
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2 Jun 2008 21:56:22 IST
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the pd due to the larger sphere is zero.
now consider the smaller sphere.
potntial at the surface is V/2 (here the distance from the centre is R)
potntial at the surface of the larger sphere is V/4 (here the distance from the centre is 2R)
so the pd is V/4.....!!!
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M sorry dude cud nt ans yesterday...
the answer is v/4 4 sure...
By definition...
V=kq/r
So for the bigger sphere...
q1=2RV/k
similarily for smaller sphere q2=Rv/2k
After sometime (-q2) charge gets induced on the outer sphere(i hope u understand it)
Therefore new charge on the bigger sphere...
q3=2RV/k- RV/2k
=3RV/2k
So the potential of the bigger sphere is V=k(3RV/2k)/2r=3/4V
The potential of the inner sphere is still V/2 as no charge is induced on its surface and hence the charge remains the same...
therefore the pot diff is (3/4-1/2)V=V/4
rate if usefull..
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