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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

```							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

```
11 years ago
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