Electrostatics

JEE Main level, single correct, medium

18 of 311

Question 18 of 311

A 10μC10\mu C charge is divided into two parts and placed at 1cm1cm distance so that the repulsive force between them is maximum. The charges of the two parts are:

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Answer: (C) 5μC,5μC5\mu C,5\mu C

Let the charges be xμC,(q−x)μCx\mu C,\left(q-x\right)\mu C, where q=10μCq=10\mu C
The force between them is
F=Kx(q−x)r2F=\dfrac{Kx(q-x)}{{r}^{2}}.
For the force to be maximum,
dFdx=0⇒dFdx=K(q−2x)r2=0⇒x=q2=5μC\dfrac{dF}{dx}=0 \Rightarrow \dfrac{dF}{dx}=\dfrac{K(q-2x)}{{r}^{2}}=0 \Rightarrow x=\dfrac{q}{2}=5\mu C

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