Question

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A charge of magnitude Q is divided into two parts q and (Q-q) such that the two parts exert maximum force on each other. Calculate the ratio Q/q.
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Answer 1

Question:-

A charge of magnitude Q is divided into two parts q and (Q-q) such that the two parts exert maximum force on each other. Calculate the ratio Q/q.

Solution:-

$\sf{Here\;F=\frac{1}{4\pi \in_{o}}\frac{q(Q-q)}{r^{2}}}$

$\sf{F\;or\;F\;to\;be\;maximum,}$

$\sf{\frac{dF}{dq}=0}$

$\sf{\frac{dF}{dq}=\frac{1}{4\pi \in_{o}r^{2}}[Q-2q]=0}$

$\sf{Q-2q=0}$

$\sf{Q=2q}$

$\sf{\frac{Q}{q}=2}$

$\sf{Q:q=2:1}$

$\boxed{\sf{Q:q=2:1}}$

Answer 2

$\huge{\mathcal{\blue{\underline{HOLA\;MATE}}}}$

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$\huge{\mathfrak{Question:-}}$

A charge of magnitude Q is divided into two parts q and (Q-q) such that the two parts exert maximum force on each other. Calculate the ratio Q/q.

$\huge{\mathfrak{Answer:-}}$

$\sf{F = \frac{1}{4\pi \inx_{o}}\times \frac{q(Q-q)}{r^{2}}}$

$\sf{\implies \frac{dF}{dq}=0\;\;\;\;\;\;\; (For\;maximum\;repulsion\;force\;derivation=0)}$

$\sf{\implies -q(2)+Q=0}$

$\sf{\implies \frac{q}{Q}=\frac{1}{2}}$

$\boxed{\sf{Q:q=2:1}}$

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