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 \(..........cm\).
Pick an option to check your answer.
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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