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Deciphering the hidden meanings

  1. Sep 25, 2011 #1
    I am new here and as common courtesy dictates, I have searched, but couldn't find, and read the requisite FAQ's and rules and didn't see anything that would either; warrant me responding in another thread, or section. It is not my intention to start off on the wrong foot, so please forgive me if I have overlooked an aspect regarding what I am trying to accomplish. My aim with this thread is to post my questions regarding terminology and equation questions. My custom study thread of sorts.

    I am currently studying the electricity and magnetism part of physics. I have found that in physics questions, that have been assigned me, there contain hidden meanings within the words they use. However, I am having a hard time figuring out on my own all the implied meanings a word is supposed to have. For instance:

    I want to make it clear I am NOT looking for the answers, those I already have, a)3.76*10^-8 and b) 4.352m, what I am looking for is what I am supposed to gather from the words they have used to describe the problem.

    So for instance, if I read "conducting sphere" I should really be thinking, "uniform charge"? Or something like that?

    The words I am trying to decode are, "conducting sphere", "charged to 680v", "relative to v=0". Are there any others I have missed?

    Thanks
     
  2. jcsd
  3. Sep 26, 2011 #2
    I think you might be making this too complicated. A conducting sphere is a sphere made of a conducting material - so yes, charge will be more or less uniform on its surface.

    A potential difference is just that - it's relative. A house on top of a hill or in a valley is house sized and its roof is as far from the ground in either case but the house at the top of the hill is higher in absolute terms. A voltage is the same - it's a potential difference. So an object held at a potential of 680v is 680v higher than something at 0v but relative to something at 10v the potential difference is 670v.

    Language needs to be used very precisely in science to avoid ambiguity. It's hard to translate unambiguous mathematical statements into unambiguous English and, I assume, other spoken or written languages.
     
  4. Sep 26, 2011 #3
    Welcome to PF, bsimpson. There's a section in PF dedicated specifically to homework
    help, but I guess your question is general enough to be answered here (I may be wrong
    though!).

    What you call 'hidden meaning' in the text is just the problem giving you all the information
    you need to solve it. Sometimes it'll be given explicitly, other times you'll have to
    deduce it from context.

    In this particular case, you're correct that a conducting sphere implies (in equilibrium)
    a constant surface charge. What this is telling you is that the problem has spherical
    symmetry (a big help!).

    Charged to 680V relative to 0V at infinity is telling you where the voltage reference
    lies (at infinity) and the voltage at the surface (relative to this reference). Thus,
    in the space surrounding the sphere (which we assume is a vacuum) the voltage
    will be maximum at the surface, and will decrease as we get further from it,
    approaching zero at infinity.

    This is enough information to solve this particular problem, and it's the same with
    any other. You need to read it carefully, and think of how the information the
    problem is giving you (and that which might not be told implicitly but must be
    deduced from context!) relates to what you know about the physics.
     
  5. Sep 26, 2011 #4

    xts

    User Avatar

    Congratulations, Bsimpson!
    You are one of those very few students, who ask such questions. My teaching experience convince me that schools to not set them, students don't ask them, examples used in textbooks (and school excercises) often are absolutely nonrealistic - then in result very few students see any correspondence between physical theories and real world.

    All physical theories always use some idealised models of reality to describe behaviour of the real objects. It is up to our knowledge, intuition, experience, common sense, and sometimes detailed analysis, to judge if the theory may be applied to some real object, and if so - how big error we may introduce.

    A 32 cm diameter conducting sphere
    It is very easy to analyse theoretically (e.g. using Gauss' law) electric field of highly symmetrical objects. Sphere is the simplest case. So theory you learn uses such idealisation. Of course, in real world there are no 'conducting spheres'. There are football balls wrapped with alufoil (it just makes about 32cm), cast iron cannonballs, and other more or less conducting more or less round objects. It is too little space here to analyse fully how good (close to real measurement if you perform the experiment) results would you get with foot ball wrepped with tinfoil instead of 32 cm sphere, but errors will be quite small, probably hard to detect them using school-grade equipment.


    is charged to 680 volts relative to v=0 at r=infinity.
    Here you have several mystic terms. First - electric potential is not an absolute value. Only the differences between potentials may be measured and have physical sense. So you always must specify the reference point (usually called V_ref=0), then when you say that your foot ball has a potential of 680V it really means, that the voltage (difference of potentials) between it and the reference is such.
    In theoretical analysis it is very convenient to chose the reference point as located far from our object, then shift it even further, and go to the limit of infinite distance. So now you have to decide what 'infinity' means in real world. You cant go with your apparatus infinitely far away. Usually, you can't go further than as to the walls of your lab room. So now you again are faced with the question: are the walls far enough to neglect their influence and use idealisation of infinity? For school-grade experiment - yes, this is justified approach.
    680V potential - it means that if you connect a voltmeter between your ball and grounded metal construction of lab walls, it would show 680V. This is the most difficult point at school-grade level, as it is very difficult to measure the potential of such ball without discharging it.

    surface charge density (sigma)?
    Some charge is stored on your ball. You may calculate it theoretically (e.g. from Gauss' law), you may measure it on several ways (but there are no ready apparata devoted to measure charge). This charge is somehow distributed over the ball. Theory says that for conductors all the charge is always stored on its surface. In real world the charge is stored within very thin (single molecule deep) layer on the surface. The charge may be distributed over the surface. In idealisation of a sphere (full symmetry) the charge is uniformly distributed over whole surface. So you may use the term of charge density: amount of charge per unit of its surface. In case of real ball wrappend with alufoil it is not necessirily true, especially if the foil is a bit wrinkly. In such case you may only approximate charge density - approximation is better if the scale is bigger. If you calculate charge density with an idealisation of sphere, you may relatively safely say that 1dm x 1dm patch of the surface contains charge Q=sigma*1dm^2. But for a bit wrinkly foil, it would be coarse error to say that 1cm x 1cm patch contains sigma*1cm^2.

    at what distance will the potential due to the sphere be only 25v?
    That's pretty interesting point. You may measure difference between potentials (although it is not so easy to measure a potential in some point of space, not occupied by any conducting object, but you may do it)
    What is interesting, is the answer you got while using theoretical idealisation: about 4m.
    This result sets limits on previously accepted experimental setup. To make this result experimentally meaningful, you must have your ball separated from any other object by significantly bigger distance than it (discussion on 'infinity' above). It means that to experimentally face with this part of your excercise, you can't perform it in the labroom. You would have to go to a big empty hall, and hang your ball in the middle: the distance from the ball to the walls, roof and ceiling must be much bigger than those 4m.
    If you perform such experiment with a ball in standard room - there will also be points of 25V potential, but full symmetry idealisation of charged sphere in empty space cannot be used to determine their position. You would have to use much more complicated theory to calculate their positions.
     
    Last edited: Sep 26, 2011
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