Let the charges be xμC,(q−x)μC, where q=10μC
The force between them is
F=r2Kx(q−x).
For the force to be maximum,
dxdF=0⇒dxdF=r2K(q−2x)=0⇒x=2q=5μC
JEE Main 2023 — Physics Electromagnetism
A 10μC charge is divided into two parts and placed at 1cm distance so that the repulsive force between them is maximum. The charges of the two parts are:
Held on 13 Apr 2023 · Verified 6 Jul 2026.
7μC,3μC
8μC,2μC
5μC,5μC
9μC,1μC
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