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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2024
  • Hard
  • MCQ

Two charges \(q\) and \(3 q\) are separated by a distance ' \(r\) ' in air. At a distance \(x\) from charge \(q\), the resultant electric field is zero. The value of \(x\) is _______.

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Answer & Solution

Correct Answer

(C) \(\frac{r}{(1+\sqrt{3})}\)

Step-by-step Solution

Detailed explanation

\(\left(\vec{E}_{\text {net }}\right)_p=0\)

\(\frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} \cdot 3 \mathrm{q}}{(\mathrm{r}-\mathrm{x})^2}\)

\((\mathrm{r}-\mathrm{x})^2=3 \mathrm{x}^2\)

\(\mathrm{r}-\mathrm{x}=\sqrt{3} \mathrm{x}\)

\(\mathrm{x}=\frac{\mathrm{r}}{\sqrt{3}+1}\)

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