Force between 2 charges increased by 16 times their distance =?

A. 4

B. 2

C. 1/8

D. 1/4

E. 8

Update:

times of original distance

times of original distance

8 Answers

  • Anonymous
    6 days ago

    Force F, between 2 charges q1 and q2 with distance r =

    F = K*q1*q2/r²

    => r = √(K*q1*q2/F)

    When Force increased by 16 times ; F becomes 16F

    r’ = √(K*q1*q2/16 F) = (1/4) * √(K*q1*q2/F) = (1/4)r

    Their distance decreased by 4 times of original distance

    Correct= D. 1/4

  • ?
    6 days ago

    D.1/4

  • Anonymous
    6 days ago

    D.1/4

  • ?
    6 days ago

    Distance indirectly proportional to root of Force.

    So Distance= root(1/16) =1/4

    D is correct.

  • Mr.Mukherjee
    6 days ago

    inversely proportional square root of the force 16 is the distance between the two charge particles, that is 1/4

  • OMPRAKASH
    6 days ago

    we know from coulomb’s law that F inversely proportional to square of distance r . So if F1 ,F2 and r1,r2 are the initial forces and distances then F1/F2=r2^/r1^ i.e 1/16=r2^/r1^ i.e r2=r1/4 so distance decreased by1/4 times. so answer D is correct

  • Hamilton
    6 days ago

    its like my third grade teacher’s stepfather’s nephew use to say, “don’t fiddle my oat bran.”

  • tom7railway
    6 days ago

    Write the question out properly please, it still doesn’t make sense.

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