 Solution

In SI units, Gauss's law for a closed surface with outward unit normal vector isSpherical symmetry and the absence of an imposed external field make the electric field radial, $\mathbf{E}=E_r(r){\mathbf{e}}_r$. For $0\le r\le a$ (with the physical radius $a>0$),Outside the charge distribution this becomes $Q(a)=8\pi K{a}^5/15$. Applying Gauss's law on spheres givesBoth formulas agree at $r=a$, and the first tends to zero at the origin. Thus the total charge is $8\pi K{a}^5/15$. Negative $K$ reverses the field direction. If the radius is zero the distribution vanishes; a negative value of $a$ gives no nonnegative-radius support under the printed piecewise inequalities and likewise gives the zero distribution.