Past exam of the physics 2 paper of the University of Oxford / 2016 / 8 / d / Solution

Set the normal electric field to zero and cancel its nonzero overall factor:Taking the power $2/3$ gives $4^{2/3}(x^2+d^2)=x^2+4d^2$, henceThe numerical value is independent of the nonzero magnitude of $Q$. Beyond this radius the signed normal electric field is positive. Thus its magnitude first falls to zero, rises again and eventually decays; it is not globally monotone.

For $x\gg d$, expanding each denominator givesThis is a dipole rather than a monopole tail. The real-plus-image charge distribution has zero total charge and dipole moment $p_z=-6Qd$; in its equatorial plane a dipole produces $E_z=-k_e{p}_z/x^3$, confirming the coefficient and sign.