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Calculate the minimum electrostatic forc...

Calculate the minimum electrostatic force between two charged particles separated at a distance of 1 cm in air.

Text Solution

Verified by Experts

As we know that, the minimum charge is an electronic
charge, i.e. `e=1.6 xx 10^(-19)C`
Also given, `r = 1cm =10^(-2)m`
So, minimum electrostatic force is the force between two electrons or between an electron and a proton.
`:. F=(kq_(1)q_(2))/(r^(2)) rArr F=(e^(2)k)/(r^(2))`
`rArr F=((1.6xx10^(-19))^(2))/((10^(-2))^(2)) xx 9xx10^(9)`
`=(2.56 xx 9 xx 10^(9) xx 10^(-38))/(10^(-4))`
`:.F_("min")=23.04 xx 10^(-25)N`
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Knowledge Check

  • When air is replaced by a dielectric medium of constant K the maximum force of attraction between two charges separated by a distance :

    A
    Decreases K times
    B
    Remains unchanged
    C
    Increases K times
    D
    Increases `K^(-2)`times
  • The force of attraction between two charges each of magnitude 3C separated by a distance 2m in air is

    A
    `20.25 xx 10^(10)N`
    B
    `20.25 xx 10^(11)N`
    C
    `20.25xx10^(8)N`
    D
    `20.25xx10^(9)N`
  • The electrostatic force between two charges Q_1 and Q_2 separated by a distance r is given by F = k =(Q_1Q_2)/(r^(2)) . The constant k

    A
    depends on the system of units only
    B
    depends on the medium between the charges only
    C
    depends on both (a ) and (b)
    D
    None of the above
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