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Question Of the Day - Feb 20 2012
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TOPIC: Question Of the Day - Feb 20 2012
#144631

Question Of the Day - Feb 20 2012 2 Years, 1 Month ago

 

A solid sphere of radius "R" is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre?

(A)R

(B)R/2

(C)R/3

(D)2R




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#144638

Re:Question Of the Day - Feb 20 2012 2 Years, 1 Month ago

 

answer should be A .... the catch here was that we had to find distance from the surface and not from the center




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#144639

Re:Question Of the Day - Feb 20 2012 2 Years, 1 Month ago

 

Nexus Are you sure i think the answer should be C.



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#144640

Re:Question Of the Day - Feb 20 2012 2 Years, 1 Month ago

 

Confirmed now!!!!

 

we know potential outside the sphere is V1 = kQ/x     where x is the distance from the center x>r

 and potential inside the sphere V1 = kQ/2r [3-x2/r2]

 for center put x=0  Vo = 3kQ/2r

condition given is V1 = Vo/2

                      kQ/x  = (3kQ/2r)/2

                         x=4r/3

 distance from the surface is  =x-r

                                         = 4r/3 -r =r/3



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#144641

Re:Question Of the Day - Feb 20 2012 2 Years, 1 Month ago

 

Please keep your eyes open!!!! And what's the catch?



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#144642

Re:Question Of the Day - Feb 20 2012 2 Years, 1 Month ago

 

here's my solution

let charge on the sphere be Q

ie potential at the centre = KQ/R

now we need half of the potential at the center ie KQ/2R

=> from center of sphere let the distance where potential is half of the center be r

KQ/r=KQ/2R

r=2R

now question asks from the surface of sphere 

ie 2R-R=R



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Last Edit: 2012/02/19 23:47 By ann_1993.
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