I have been trying to understand the last step of this derivation. Consider a sphere made up of charge +q. Let R be the radius of the sphere and O, its center.
A point P lies inside the sphere of charge. In such a case, Gaussian surface is a spherical shell,whose center is O and radius is r (=OP). If q′ is charge enclosed by Gaussian surface,then
E×4πr2=q′/ϵ where ϵ= absolute permittivity of free space.
E=14πϵ×(q′/r2) for $(r
q′=q43πR3×43πr3=qr3R3
Answer
I presume your problem is the calculation of q′=r3R3q.
This is perhaps easier to explain by splitting the calculation in two steps. The solid ball of charge is supposed to be homogeneous, so it has a charge density ρ=total chargetotal volume=q4π3R3. The smaller sphere has volume Vr=4π3r3, and therefore has charge q′=ρVr=q4π3R34π3r3=r3R3q.
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