Question 21 of 311

A point charge ${q}_{1}=4{q}_{0}$ is placed at origin. Another point charge ${q}_{2}=-{q}_{0}$ is placed at $x=12cm$. Charge of proton is ${q}_{0}$. The proton is placed on $x$-axis so that the electrostatic force on the proton in zero. In this situation, the position of the proton from the origin is ______ $cm$.

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Answer: 24

Let at point $P$ net electric field is zero.
As direction of electric field due to both charges at point $P$ is opposite, equating magnitude we get
$\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{4{q}_{0}}{{x}^{2}}=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{{q}_{0}}{{\left(x-12\right)}^{2}} \Rightarrow 2\left(x-12\right)=x \Rightarrow x=24cm$
Distance of proton from origin $=24cm$.