The electric field strengh in region is given as E=Xi+Yj / X^2+Y^2. Find the net charge inside a sphere of radius 'a' with its center at origin. (a=2M)write value of 'k' if charge is k epsulon.

							
consider the sphere as a gaussian surface then,
    § E . ds  = Q in /£ ..........................(1)
 E   =  Xi+Yj / X^2+Y^2   at any point (x,y) on the surface of the sphere ,
 area vector is always parallel to the surface, so in the case of the sphere it 
will be in the direction of radius vector ,
so,  at any point (x,y)  the area vector  Xi+Yj  /  (X^2+Y^2) 1/2  ds ,
                 where  Xi+Yj  /  (X^2+Y^2)  is the unit vector in the direction of area vector  
so the l.h.s of eq (1),
      § E . ds   =
                       = §(Xi+Yj / X^2+Y^2) .{ Xi+Yj  /  (X^2+Y^2) 1/2 }  ds 
                    =  § (X^2+Y^2) / (X^2+Y^2) 3/2 ds
                    = §1/ (X^2+Y^2) 1/2  ds
 
( for a sphere  (X^2+Y^2) 1/2  =  radius = 2M)
                   = 1/2M  §ds
         (   total surfase area for a sphere =  4 pi r^2
                                                                  = 4 pi * 4 M^2)
                        =  1/2M  * 16 *pi * M^2
                     =  8 pi m^3 =   r.hs.
so 
  8 pi m^3  =  Q in /£ 
 Q = 8 pi m^3 £
           k= 8 pi m^3