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JEE Mains · Physics · STD 12 - 1. Electric charges and fields

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

  1. A \(24\)
  2. B \(23\)
  3. C \(22\)
  4. D \(20\)
Verified Solution

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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