Image charge for a grounded conducting sphere ($q^{\prime }=-aq/R,{\mathbf{p}}^{\prime }=a^2\mathbf{p}/R^2$)

For a charge $q$ at $\mathbf{p}$, $R=|\mathbf{p}|>a$, outside a grounded sphere of radius $a$, the displayed image gives zero potential on the sphere and decay at infinity. On $|\mathbf{x}|=a$, one has $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$, proving boundary cancellation. The induced force is inward, of magnitude $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$, from Coulomb's law applied to the image field. The self-field is excluded.

Place equal positive point charges at distances $d$ on opposite sides of a grounded conducting sphere of radius $a$, and set $x=d/a>1$. Each physical charge has an image $-qa/d$ at distance $a^2/d$ from the centre on its own side. The repulsion from the other physical charge balances attraction to both images precisely when the displayed equation holds. At large $x$, the two sides behave as $1/(4x^2)$ and $2/x^3$, explaining the approximate root $x\simeq 8$. A constant nonzero sphere potential adds a central image and changes the force balance.

A real charge $q$ at $d>a$ outside a spherical conductor of radius $a$ has the usual grounded-sphere image charge $q^{\prime }=-aq/d$ at $x^{\prime }=a^2/d$. Adding a central image $q_0=-q^{\prime }$ makes the total image charge zero, representing an isolated neutral conductor. The off-centre pair has zero potential on $r=a$; the central image therefore sets the constant conductor potential to $q/(4\pi {\epsilon }_{0}d)$. Its induced electric dipole moment is $p_x=q^{\prime }{x}^{\prime }=-q{a}^3/d^2$. Keeping the applied central field ${\mathbf{E}}_0=-q\hat{\mathbf{x}}/(4\pi {\epsilon }_0{d}^2)$ fixed gives $\mathbf{p}=4\pi {\epsilon }_0{a}^3{\mathbf{E}}_0$. As $d\to \infty$, the external field becomes uniform near the sphere; its divergent constant potential can be removed without affecting the electric field.