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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2025
  • Easy
  • MCQ

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 :

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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation


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 \(\left(\frac{\mathrm{d}}{4}, 0,0\right)\)
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