Question

A solid non-conducting sphere of radius R has a nonuniform charge    distribution of volume charge density ρ = rρ_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_sR³ and

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

       E = (Q r² / 4π ε₀R⁴)

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 R² ) dR ]

= p_s x pie x 4 x R integration of [ r . dR ]

= p_s x pie x 4 x R x ( R² / 2 )

= 2 x pie x p_s x R³ 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 r² ) = pV / e_o

so the electric field in the gauss surface

E = pV / ( 4 x pie x e_o x r² )

   = pr / 3e_o  r^{^}

here p = Q / V_total

so E = Qr / ( 4/3 x pie x R³ ) x 3 x e₀

   E = Qr / ( 4 x pie x e_o x R³ ) r^{^}

here vector r^{^} = r / R

so the magnitude of electric field

E = Qr² / ( 4 x pie x e_o x R⁴ )