JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 :
- A \(9\,\mu C , 1\,\mu C\)
- B \(5\,\mu C , 5\,\mu C\)
- C \(7\,\mu C , 3\,\mu C\)
- D \(8\,\mu C , 2\,\mu C\)
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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