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  • A charge of is to be divided into two. The distance between the two divided charges

A charge of 4 \mu \mathrm{C} is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be :
 

Option: 1

1 \mu \mathrm{C}$ and $3 \mu \mathrm{C}


Option: 2

2 \mu \mathrm{C}$ and $2 \mu \mathrm{C}


Option: 3

\mathrm{0 \: and\: 4 \mu \mathrm{C}}


Option: 4

\mathrm{1.5 \mu \mathrm{C} \: and \: 2.5 \mu \mathrm{C}}


Answers (1)

best_answer

\mathrm{F=\frac{KQ\left ( 4-Q \right )}{r^{2}}}

For maximum force,

\mathrm{\frac{dF}{dr}=0}

\mathrm{Q=\frac{4}{2}=2\mu c}

The magnitude of the divided charges so that the force between them is maximum,will be \mathrm{2\mu c\: and\: 2\mu c}

Hence 2 is correct option.

Posted by

Suraj Bhandari

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