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A certain charge 2Q is divided at first ...

A certain charge 2Q is divided at first into two parts `q_(1)` and `q_(2)`. Later the charges are placed at a certain distance. If the force of interaction between two charges is maximum then `Q/(q_(1))` = ________.

A

1

B

4

C

0.5

D

2

Text Solution

Verified by Experts

The correct Answer is:
A
(a): Let r be the distance between `q_(1)` and `q_(2)`.
According to coulomb's law, the force between them is
`F=(1)/(4piepsilon_(0)) (q_(1)q_(2))/(r^(2)) = 1/(4piepsilon_(0)) (q_(1)(2Q-q_(1)))/(r^(2)) ( :' q_(1)+q_(2) = 2Q)` For F to be maximum `(dF)/(dq_(1)) = 0` as Q and r are constant.
`:. d/(dq_(1)) [(1)/(4piepsilon_(0)r^(2)) (2Qq_(1)-q_(1)^(2))]=0`
`2Q - 2q_(1) = 0` , Hence, `Q/(q_(1))` = 1
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