plzz help me with this i need to know 100% correct answer?

I have two equal positive charges placed 38 cm apart. They are moved so the force between them decreases by a factor of 6 times the original force. What is the distance between them?

5 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    that depends you should explain your question better

  • Ian I
    Lv 4
    1 decade ago

    Electrostatic repulsive forces obey the inverse square law. This means that if they experience a force F at a distance R, then the magnitude of the force is given by F = N/(R^2).

    So, in this case, you start off with force F = N/(38^2) = N/1444, where N is a constant.

    At what distance 'r' will F be reduced to (1/6)F?

    N is constant, but F is REDUCED by a factor of 6. This means that R^2 must have INCREASED by a factor of 6.

    So, 1444 x 6 = 8664 = new R^2

    So, new R is sqrroot 8664 = 93 cm.

  • Anonymous
    1 decade ago

    No such distance exists.

    Moving the cahrges apart to infinity will decrease the force between them to zero. This is a reduction of 1 times the original force. There can be no greater reduction.

  • Anonymous
    1 decade ago

    94.28 cm

    F1=kQ²/r²

    F2=kQ²/(r+x)²

    (r+x)²=6r²

    r²+2rx+x²=6r²

    if r=38, then:

    1444+72x+x²=8664

    x²+72x-7220=0

    x=(-72+sqrt(72²+4*7220))/2 = 56.28

    So the new distance is r+x = 38+56.28 = 94.28

  • How do you think about the answers? You can sign in to vote the answer.
  • 1 decade ago

    ozo has the right approach here

Still have questions? Get your answers by asking now.