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Homework Help: Electric force between two spheres?

  1. May 19, 2004 #1
    Two identical small spheres possessing charges q1 and q2 are seperated by distance r. Which charge would produce the greatest inecreas in the electric force between the tow spheres?

    1. double charge q1
    2. double r
    3. double r and q1
    4. double r, q1, and q2

    I'm pretty sure it's #1
  2. jcsd
  3. May 19, 2004 #2


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    Since force goes as product of charges and inverse square of distance,

    #1 doubles force
    #2 cuts force to 1/4 of what it was
    #3 cuts force to 1/2 of what it was
    #4 leaves force the same

    So you are correct in answering #1.
  4. May 19, 2004 #3
    The resistance of a wire at constant temperature depends on the wire's...length and cross sectional area. Does the type of meteal come into play?
  5. May 19, 2004 #4
    Of course it does.

    Look up resistivity.
  6. May 19, 2004 #5
    ohhh, right R=resistivity*L/A

    In a parallel connection of two resistors, if the resistance of the resistors were increased, the current would decrease right? According to R=V/I
  7. May 19, 2004 #6
    You get a *qualified* yes there.

    Yes, if you increase the resistance the current decreases. But remember that if the resistors are in parallel you can't simply add them together, right?
  8. May 19, 2004 #7
    In a parallel, if the total resistance decreases when the resistors increase in resistance dure to 1/Rt=1/R1+1/R2...

    So, if resistance in the resistors were to increase in a parallel, then really, the total resistance would decrease making the current increase??

    I'm confused now
  9. May 19, 2004 #8
    2 resistors in parallel: r1 = 4; r2 = 5
    [tex]\frac{1}{4} + \frac{1}{5} = \frac{9}{20} \textrm{giving} R_T = \frac{20}{9}[/tex]

    increase: now r2=5
    [tex]\frac{1}{5} + \frac{1}{5} = \frac{2}{5} \textrm{giving} R_T = \frac{5}{2}[/tex]

    5/2 is greater than 20/9


    Effective total resistance has increased, so the current will decrease.
  10. May 19, 2004 #9


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    Increasing R1 would decrease 1/R1. Increasing R2 would decrease 1/R2. Doing either one of those will decrease the right side, and therefore the left side must also decrease. But that means the reciprocal of the left side would increase, and the reciprocal is in fact the total resistance, so total resistance goes up.
  11. May 19, 2004 #10


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    Gnome beat me to the answer.
  12. May 19, 2004 #11
    Okay, but the more resistors you add, the less the total resistance right?
  13. May 19, 2004 #12


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    That is true, as long as you add them in parallel with the other ones.
  14. May 19, 2004 #13
    thanks guys, that cleared up a lot of things
  15. May 19, 2004 #14
    Urban, do you realize the difference between increasing the resistance of an existing resistor vs. adding an additional resistor in parallel with the others?

    In the first instance, you are further restricting the flow of current; the combined effective resistance is increased.

    In the second, you are actually adding an additional pathway for the current to flow through, so the combined effective resistance is decreased.
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