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JEE MAINS 2024
31.01.24 S1
Question
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
Select the correct option:
A
$\frac{(1+\sqrt{3})}{r}$
B
$\frac{r}{3(1+\sqrt{3})}$
C
$\frac{r}{(1+\sqrt{3})}$
D
$r(1+\sqrt{3})$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. $$ \begin{aligned} & \left(\vec{E}_{n e t}\right)_p=0 \\ & \frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} \cdot 3 \mathrm{q}}{(\mathrm{r}-\mathrm{x})^2} \\ & (\mathrm{r}-\mathrm{x})^2=3 \mathrm{x}^2 \\ & \mathrm{r}-\mathrm{x}=\sqrt{3} \mathrm{x} \\ & x=\frac{r}{\sqrt{3}+1} \end{aligned} $$
Question Tags
JEE Main
Physics
Easy
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