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

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Two point charges $$Q$$ and $$Q^\prime$$ are located at positions given by $$\vec{a}, \vec{b}, (a\ne b)$$. Find the conditions on $$Q, Q^\prime, a, b$$  so that the potential on the surface of a sphere of radius $$R$$ with center at the origin may be zero.

Note the result of this problem is used in solution for electric potential of a grounded conducting sphere in presence of a point charge by the method of images.

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} Download: em-ymp-04001.pdf (34.42 KB)

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