Question 65 of 311

A point charge +q is placed at the origin. A second point charge +9 q is placed at $(\mathrm{d}, 0,0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is :

A$(4 \mathrm{~d} / 3,0,0)$B$(\mathrm{d} / 4,0,0)$C$(3 \mathrm{~d} / 4,0,0)$D$(\mathrm{d} / 3,0,0)$

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Answer: (B) $(\mathrm{d} / 4,0,0)$

Let $\mathrm{E}_{\mathrm{p}}=0$
$\begin{aligned}
& \therefore \frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} 9 \mathrm{q}}{(\mathrm{~d}-\mathrm{x})^2} \\
& \Rightarrow \frac{\mathrm{~d}-\mathrm{x}}{\mathrm{x}}=3 \Rightarrow \mathrm{x}=\frac{\mathrm{d}}{4}
\end{aligned}$
$\therefore$ co-ordinate of P is $\left(\frac{\mathrm{d}}{4}, 0,0\right)$