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For a given separation between the two p...

For a given separation between the two point charges, the force between them is `64xx10^(-4)N`. If the distance is decreased by 0.25 m then the force between them is `100xx10^(-4)N`.Find the initial separation between the charges.

Text Solution

Verified by Experts

`(F')/F=(d/(d'))^(2)` where `d'=d-0.25`
i.e `(d/(-0.25))^(2)=((100xx10^(-4))/(64xx10^(-4)))`
or `d/((d-0.25))=5/4,4d=5d-1.25`
`:.d=1.25m`
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