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Two point charges placed at a distance `r` in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant `K` is

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Force between two point charges in air at certain distance is 200N. What will force when they are kept in a medium of dielectric constant 4, at the same distance?

Force between two point charges in air at certain distance is 200N . What will force when they are kept in a medium of dielectric constant 4 , at the same distance?

Knowledge Check

  • Two point charges placed at a certain distance r in air exert a force F on each other. Then the distance r at which these charges will exert the same force in a medium of dielectric constnat K is given by

    A
    r
    B
    `r/K`
    C
    `r/(sqrt(K))`
    D
    `rsqrt(K)`

  • Two point charges placed at a distance r in air exert a force f on each other. The value of distance R at which they experience force 4F when placed in a medium of dielectric constant K = 16 is :

    A
    r
    B
    r/4
    C
    r/8
    D
    2r

  • The force of attraction between two charges at distance r, when the air is replaced by a medium of dielectric constant k.

    A
    Increases `k^(2)` times
    B
    decreases K times
    C
    remains uncharged
    D
    increases K times

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    Two point charges placed at a distance of 0.20m in air repel each other with a certain force.When a dielectric slab of thickness 0.08m is introduced in between the charges,the force of interaction is half of its previous value.The value of dielectric constant of slab is ,,,..

    Two similar point charges q_1 and q_2 are placed at a distance r apart in the air. The force between them is F_1 . A dielectric slab of thickness t(ltr) and dielectric constant K is placed between the charges . Then the force between the same charge . Then the fore between the same charges is F_2 . The ratio is

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