Question 73 of 311

Two charges $q$ and $3q$ 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$\dfrac{(1+\sqrt{3})}{r}$B$\dfrac{r}{3(1+\sqrt{3})}$C$\dfrac{r}{(1+\sqrt{3})}$D$r(1+\sqrt{3})$

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Answer: (C) $\dfrac{r}{(1+\sqrt{3})}$

Given: ${\left({\vec{E}}_{net}\right)}_{P}=0$
$\Rightarrow \dfrac{kq}{{x}^{2}}=\dfrac{k\times 3q}{{\left(r-x\right)}^{2}}$
$\Rightarrow {\left(r-x\right)}^{2}=3{x}^{2}$
$\Rightarrow r-x=\sqrt{3}x$
$\Rightarrow x=\dfrac{r}{\sqrt{3}+1}$

JEE Main Physics practice question on Electrostatics (Electric Field and Electric Field Lines), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.