[YMP/EM-04001] Grounded Conducting Sphere and a Point Charge

The potential due to the two charges at point \(\vec{r}\) is given by \[ \Phi(\vec{r}) = \frac{1}{4\pi\epsilon_0}\frac{Q}{|\vec{r}-\vec{a}|} + \frac{1}{4\pi\epsilon_0}\frac{Q^\prime}{|\vec{r}-\vec{b}|}\] The condition that the potential at an arbitrary point P, position vector \(= R \hat{n}\), on the sphere be zero is

\begin{eqnarray}
 \frac{1}{4\pi\epsilon_0}\frac{Q}{|\vec{r}-\vec{a}|} +
\frac{1}{4\pi\epsilon_0}\frac{Q^\prime}{|\vec{r}-\vec{b}|}&=&0 \\
  Q^\prime {|R\vec{n}-\vec{a}|} +
    Q{|R\vec{n}-\vec{b}|}&=&0\label{EQ01}
\end{eqnarray}

By symmetry the centre of the sphere and the positions of the two point charges must be in a straight line. Thus vectors \(\vec{a}\) and \(\vec(b)\) must be parallel and we write \[ \vec{a} = a\hat{m}, \qquad \vec{b} = b \hat{m}.\] Squaring the relation \eqref{EQ01} we demand that

\begin{eqnarray}
  Q^{\prime 2}\{ R^2+ b^2 -2Rb\, \hat{m}\cdot\hat{n}\}  =  Q^2\{ R^2+ a^2 -2Ra\,
\hat{m}\cdot\hat{n}\}
\end{eqnarray}

The above relation must hold for all \(\hat{m} \cdot \hat{n}\). This gives two relations

\begin{eqnarray}
  Q^{\prime 2} b  &=&  Q^2 a \label{EQ02}\\
    Q^{\prime 2}\{ R^2+ b^2 \}  &=& Q^2\{ R^2+ a^2 \}\label{EQ03}
\end{eqnarray}

Substituting \(Q^{\prime2}=Q^2(a/b)\), from \eqRef{EQ02}, in \eqref{EQ03}, gives

\begin{eqnarray}
  &&  Q^2 \frac{a}{b}(R^2+b^2)= Q^2 (R^2+a^2)\\
\Rightarrow &&   aR^2+ab^2 =bR^2 + ba^2\\
\Rightarrow && (a-b)R^2 - ab(a-b) =0 \\
\Rightarrow && R^2 = ab \\
\end{eqnarray}

It is obvious that the sign of \(Q\) and \(Q^\prime\) must be opposite. Hence

\begin{eqnarray}
  Q^{\prime} &=& - Q \sqrt{a/b}
\end{eqnarray}

Thus we have the answer \begin{equation} \boxed{ Q^\prime = - \frac{Q a^2}{R}, \qquad   b =  \frac{R^2}{a}.} \end{equation}