# 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.

Pratham Ashish
17 Points
14 years ago

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