A 10,μ C charge is divided into two parts and placed at 1,cm distance so that repulsive force betwe

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1. Electric Charges and Fields

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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

$8\,\mu C , 2\,\mu C$

(JEE MAIN-2023)

Solution

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

$\frac{d F}{d x}=\frac{K}{r^2}(q-2 x)=0$

$x =\frac{ q }{2}$

Standard 12

Physics

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