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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: gravitational field
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feynmann (2189)

Forum Expert Blazing goIITian

Olaaa!! Perrrfect answer. 395  [502 rates]

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total posts: 806    
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It's simple
 
See  E = GM/ r^3  r
            = 4/3 pi  r .......................... ( 1 )
( M is the mass enclosed )
      fist suppose that the cavity was not there . And we are interested in finding E at the point p from the centre of the cavity . The point that we are considering is at r from the centre of the big sphere .
 
So we can write
                               r = p + l ........................ ( 2 ) 
 now we can write that
 
          E ( due to solid sphere ) = E ( due to sphere with its cavity ) + E ( due to solid cavity )
 
so,  E ( due to sphere with its cavity )
 
 =            E ( due to solid sphere ) - E ( due to solid cavity ) .............. ( 3 )
 
Now using ( 1 ) & ( 3 )
 
    we have reqd E = 4/3 pi  ( r - p )
                             = 4/3 pi  l         ( from ( 2 ) )
which is constant throughout the cavity 
 this reply: 15 points  (with Olaaa!! Perrrfect answer.   in 3 votes )   [?]
 
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