Results 1 to 10 of 10

Math Help - tiny spheres conducting

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    27

    Unhappy tiny spheres conducting

    Two tiny conducting spheres are identical and carry charges of -21.6 mC and +51.4 mC. They are separated by a distance of 2.43 cm. What is the magnitude of the force that each sphere experiences?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by lovinhockey26 View Post
    Two tiny conducting spheres are identical and carry charges of -21.6 mC and +51.4 mC. They are separated by a distance of 2.43 cm. What is the magnitude of the force that each sphere experiences?
    Either
    F = \frac{kQ_1Q_2}{r^2}
    or
    F = \frac{Q_1Q_2}{4 \pi \epsilon _0 r^2}
    depending on how your book treats it.

    (Note: r will be the distance between the centers of the two spheres. The problem did not specify this.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
    27

    still lost

    okay...
    so using that formula i did the following calculations....

    -.0000216 x .0000514/5.9049


    which gave me a final answer of -1.88E-10 N

    which is apparently the wrong answer... what am I missing here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by lovinhockey26 View Post
    Two tiny conducting spheres are identical and carry charges of -21.6 mC and +51.4 mC. They are separated by a distance of 2.43 cm. What is the magnitude of the force that each sphere experiences?
    Quote Originally Posted by lovinhockey26 View Post
    okay...
    so using that formula i did the following calculations....

    -.0000216 x .0000514/5.9049


    which gave me a final answer of -1.88E-10 N

    which is apparently the wrong answer... what am I missing here?
    The "m" in -21.6 mC is "milli" or "micro?" (It reads as "milli" so I don't know if you have that right or not.)

    Also the radius is in cm, not m, so the radius squared is (2.43 \times 10^{-2}~m)^2 = 0.00059049~m^2

    Finally you forgot the constant out at the front: k = 9 \times 10^9 Nm^2/C^2.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2008
    Posts
    27
    it is supposed to be micro coloumbs
    in my problem it displays the symbol for mu, but when i copy/paste the mu is lost and simply becomes the letter m... hence..its supposed to be 10^-6, or at least from my knowledge that should be right....
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2008
    Posts
    27

    hmm

    when i multiplied the constant by the remaining number (after fixing my cm to m) I came up with 15246...
    which I'm fairly confident is incorrect....

    I don't know why I keep getting the wrong numbers...I am also unsure of how to label my answer..that is if I ever figure it out...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by lovinhockey26 View Post
    when i multiplied the constant by the remaining number (after fixing my cm to m) I came up with 15246...
    which I'm fairly confident is incorrect....

    I don't know why I keep getting the wrong numbers...I am also unsure of how to label my answer..that is if I ever figure it out...
    I don't know why your calculator isn't working either. This is just "plug-n-chug" stuff.

    F = \frac{(9 \times 10^9)(-21.6 \times 10^{-6})(51.4 \times 10^{-6})}{(2.43 \times 10^{-2})^2} = -194400

    As to the units:
    k has a unit of \frac{Nm^2}{C^2}

    each charge has a unit of C

    and r has a unit of m.

    Thus F has units of
    \frac{\frac{Nm^2}{C^2} \cdot C \cdot C}{m^2} = N
    as we expect, since F is a force.

    -Dan
    Last edited by topsquark; February 26th 2008 at 07:19 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Feb 2008
    Posts
    27

    :(

    I tried the exact way that you have it posted here...

    i tried submitting the answer and it doesn't like my answer...

    I'm down to my last try and I don't understand what could possibly be wrong...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Feb 2008
    Posts
    27

    maybe

    okay,
    so I tried it yet again... just calculating numbers... I used your setup that you have posted here....
    I came up with -16968.81 N

    Is it possible that this could be the answer?? Or am I still missing something somewhere with my calculations?

    I'm not going to plug in the answer to the site until I'm more confident, because I really can't afford to lose the points...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by lovinhockey26 View Post
    okay,
    so I tried it yet again... just calculating numbers... I used your setup that you have posted here....
    I came up with -16968.81 N

    Is it possible that this could be the answer?? Or am I still missing something somewhere with my calculations?

    I'm not going to plug in the answer to the site until I'm more confident, because I really can't afford to lose the points...
    I'm getting -16921.8 N. I'm not sure why there is a difference between our answers.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Conducting Math Research
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: August 11th 2011, 04:52 AM
  2. Conducting a Better Statistics Study
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 20th 2010, 12:56 PM
  3. Three tiny questions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 9th 2009, 03:45 AM
  4. Non conducting disc?
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: August 8th 2009, 12:11 PM
  5. Replies: 9
    Last Post: June 27th 2008, 11:11 PM

Search Tags


/mathhelpforum @mathhelpforum