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Electric charge and fields

  1. Jul 7, 2003 #1


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    I have some missed test questions that I still don't understand. Can anyone please explain these.

    Two chrged objects atrract each other with a force F. What happens to the F between them if one is doubled, the other is tripled and the seperation distance between their centers is reduced to 1/4 its original value? the force is now equal to


    2)What is the charge on 1kg of protons?

    9.6 x 10^7 C

    3)A copper penny has mass 3g. A total of 4x10^12 electrons are transfored from one nuetral penny to another. if the electrostatic force of attraction between the pennies is equal to the weight of a penny, what is the seperation between them?


    4) How can a negatively charged rod charge an electroscope negatively? Is it by conduction because its not induction.

    can anyone gimme their answer and how they derived it since I missed it I provided the remaining test answers to help you choose from. Please any help is appreciated as always.

    Dx :wink:
  2. jcsd
  3. Jul 7, 2003 #2

    Tom Mattson

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    As you know, I don't like posting complete solutions here, especially when the poster hasn't shown any work. It doesn't teach you anything when someone shows you how to solve a problem. So, I am going to give you hints so that you can figure them out for yourself.

    First, note that all 3 changes serve to increase the force, so you can eliminate choice #2 immediately.

    Second, take the expression for Coulomb's law:


    Now let q1-->2q1
    and q2-->3q2
    and r-->(1/4)r

    Make the substitutions in Coulomb's law and answer the question.

    Easy as pie.

    You need the mass and charge of a proton, and from there it's a simple unit conversion.

    They give you the magnitude of the electric force. It equals mpennyg. They also give you the charge on each penny. Use Coulomb's law:


    The only unknown is r, and that is what they are asking for.

    There's no hint I can think of that doesn't give it away. You'll have to do the research on it.
  4. Jul 7, 2003 #3
    I don't understand the first question.

    Is it asking: What is the force between the two chrages if the charge of one particle is double and the other trippled and their distance of seperation cut by a quarter.

    Set it up:

    F = k * q^2/r^2
    Ff = k * 6q^2/(1/4r)^2

    Mentor Edit: Rest of solution removed.

    I apprciate anyone who helps out in this forum, but let's stick to giving hints. A few complete solutions to problems have been posted lately, and I'm cracking down!
    Last edited by a moderator: Jul 7, 2003
  5. Jul 7, 2003 #4


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    You did it exactly as i did, I got 24F and it was marked wrong on my test.

    Please understand I am serious about cheating also and don't believe in it fellas, plz understand your help is appreciated and ALL I AM ASKING is how you solved it. Thats it! Besides I can't learn if I don't know whats going on and I hate typing all my work just to get a few sentences of "HINTS". So ty to understand im not about cheating I just want to understand why I missed those questions and I appreciate everyones help. Tom, you always do a great job of explaining, i thank you, sir.
    Dx :wink:
    Last edited by a moderator: Jul 7, 2003
  6. Jul 7, 2003 #5

    Tom Mattson

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    Cheating is not the issue. The issue is that, if we simply hand the solution and an explanation to you, you aren't learning anything. You have to figure these things out for yourself, and if you post your work we can help you with that (This is the Homework Help forum, after all).

    Post your work, and we will tell you where you went wrong.
  7. Jul 7, 2003 #6


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    I see! But thats sure alot of typing I must do.:wink:
    I'll comply from this day forward.
  8. Jul 7, 2003 #7

    What's 6/(1/4)2?


    Opps! Edit
    Last edited: Jul 7, 2003
  9. Jul 7, 2003 #8
    Ummmm: Sorry, I wasn't really thinking about that. I usually don't post whole solutions.
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