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 \(..........cm\).
- A \(24\)
- B \(23\)
- C \(22\)
- D \(20\)
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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