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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. 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
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
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