JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2023
- Hard
- MCQ
A \(10\,\mu C\) charge is divided into two parts and placed at \(1\,cm\) distance so that the repulsive force between them is maximum. The charges of the two parts are :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(5\,\mu C , 5\,\mu C\)
Step-by-step Solution
Detailed explanation
Divide \(q =10\,\mu C\) into two parts \(x\) and \(q - x\)
\(F =\frac{K x(q-x)}{r^2}\)
For \(F\) to be maximum
\(\frac{d F}{d x}=\frac{K}{r^2}(q-2 x)=0\)
\(x =\frac{ q }{2}\)
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