Question

A solid non-conducting sphere of radius *R* has a
nonuniform charge distribution of volume charge
density ρ = *r*ρ_{s}/*R*, where ρ_{s}
is a constant and *r* is the distance from the
centre of the sphere.

Show that:

(a) the total charge on the sphere is Q = π
ρ_{s}*R*^{3} and

(b) the magnitude of the electric field inside the sphere is given by the equation

E = (Q *r*^{2} /
4π ε_{0}*R*^{4})

Answer #1

here radius of sphere = R variable charge density p = r
p_{s} / R

as we know that the volume harge density

p = dQ / dV

so the total charge

Q = integration of [ p . dV ]

= integration of [ p . A . dR ]

= integration of [ ( r.p_{s} / R ) ( 4 x pie x
R^{2} ) dR ]

= p_{s} x pie x 4 x R integration of [ r . dR ]

= p_{s} x pie x 4 x R x ( R^{2} / 2 )

= 2 x pie x p_{s} x R^{3} Cb Ans

(b) as we know that

first let a gauss surface of radius r

then flux fie = EA

and the flux is also given as fie = Q / e_{o}

so EA = Q / e_{o}

E( 4 x pie x r^{2} ) = pV / e_{o}

so the electric field in the gauss surface

E = pV / ( 4 x pie x
e_{o} x r^{2} )

= pr /
3e_{o} r^{^}

here p = Q /
V_{total}

so E = Qr / ( 4/3 x pie x
R^{3} ) x 3 x e_{0}

E = Qr / ( 4 x pie x
e_{o} x R^{3} ) r^{^}

here vector r^{^} = r / R

so the magnitude of electric field

E = Qr^{2} / ( 4 x pie x e_{o} x R^{4}
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