In Ex. 3.9, we obtained the potential of a spherical shell with surface charge σ(θ) = k cos θ. In Prob. 3.30, you found that the field is pure dipole outside; it’s uniform inside (Eq. 3.86). Show that the limit R → 0 reproduces the delta function term in Eq. 3.106.
V(r, \theta)=\frac{k}{3 \epsilon_{0}} r \cos \theta \quad(r \leq R) (3.86)
E _{\text {dip }}( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ]-\frac{1}{3 \epsilon_{0}} p \delta^{3}( r ) (3.106)
We need to show that the field inside the sphere approaches a delta-function with the right coe!cient (Eq. 3.106) in the limit as R → 0. From Eq. 3.86, the potential inside is
V=\frac{k}{3 \epsilon_{0}} r \cos \theta=\frac{k}{3 \epsilon_{0}} z, \quad \text { so } \quad E =-\nabla V=-\frac{k}{3 \epsilon_{0}} \hat{ z } .
From Prob. 3.30, the dipole moment of this configuration is p =\left(4 \pi R^{3} k / 3\right) \hat{ z }, \text { so } k \hat{ z }=3 p /\left(4 \pi R^{3}\right) , and hence the field inside is
E =-\frac{1}{4 \pi \epsilon_{0} R^{3}} p .
Clearly E → ∞ as R → 0 (if p is held constant); its volume integral is
\int E d \tau=-\frac{1}{4 \pi \epsilon_{0} R^{3}} p \frac{4}{3} \pi R^{3}=-\frac{1}{3 \epsilon_{0}} p ,
which matches the delta-function term in Eq. 3.106.