A point charge q₁=4 q_0 is placed at origin. Another point charge q₂=-q_0 is placed at x =12,cm

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1. Electric Charges and Fields

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A point charge $q_1=4 q_0$ is placed at origin. Another point charge $q_2=-q_0$ is placed at $x =12\,cm$. 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 $..........cm$.

A

$24$

B

$23$

C

$22$

D

$20$

(JEE MAIN-2023)

Solution

$\frac{q_0}{x^2}=\frac{4 q_0}{(x+12)^2}$

$x+12=2 x$

$x=12$

Distance from origin $= x +12=24\,cm$.

Standard 12

Physics

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