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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2023
  • Medium
  • Numerical

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 \(..........cm\).

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Answer & Solution

Correct Answer

(A) \(24\)

Step-by-step Solution

Detailed explanation

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

\(x+12=2 x\)

\(x=12\)

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

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