JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 _______.
- A \(\frac{(1+\sqrt{3})}{r}\)
- B \(\frac{\mathrm{r}}{3(1+\sqrt{3})}\)
- C \(\frac{r}{(1+\sqrt{3})}\)
- D \(r(1+\sqrt{3})\)
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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