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Grade: 12

                        

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.

11 years ago

Answers : (1)

Pratham Ashish
17 Points
							

consider the sphere as a gaussian surface then,

    § E . ds  = Q in /£ ..........................(1)

   =  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

 

 

11 years ago
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