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Intensity of test charge.

  1. Mar 31, 2009 #1
    The intensity of the test charge should be very less cause it will disrupt the field.

    Question is will it matter even if it disrupts the field?

    Consider 2 point charges q1 and q0 (take this as the test charge), as the intensity of 2 increases, so does the magnitude of force that applies on the 2, what does this mean?...it means that that though the field of the 2 is getting disturbed, the attractive/repulsive force are not; so the attractive/repulsive forces should be independent of the degree of disturbance in the field which they mutually cause.

    So why does the test charge need to be small, it'll give the same result. Lets analyze what I've said by the definition of E.F -

    E = F/q0

    Considering a constant E.F, as q0 increases, so will F ( Coulomb's law) and so the ratio will remain constant..............so the value of q0 should not matter at all!


    In fact how do we even come to the conclusion that under addition of new charges within the field of another source charge, the E.F of that charge gets disturbed? Cause if a test charge is put in this scenario, the force on it will be altered?...well it's altered cause of the other charge, that is the force on this test charge will be a resultant.
     
  2. jcsd
  3. Apr 5, 2009 #2
    Can someone pls help me with this?...exams close.
     
  4. Apr 5, 2009 #3

    jtbell

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    Staff: Mentor

    Hint: F = ma. :wink:
     
  5. Apr 5, 2009 #4
    f(E, r) and m does not matter.

    Also a(f).

    How can that help?

    Sorry I'm too dumb for this.
     
  6. Apr 5, 2009 #5
    Consider that F=ma acts over time. Consider th dynamics. What would disrupt the system by a large test charge that should be uneffected by an infintesimal test charge? Hope this helps.
     
  7. Apr 5, 2009 #6
    You need to carefully determine what question you are asking yourself. Are you asking if a large test charge can be used to measure an existing electric field?

    Consider what happens if you bring a test charge into an electric field generated by one or more other charges. The force on the test charge is proportional to it's own charge as it interacts in the preexisting field. So field can be measured as force on the test charge divided by its charge. However, if the test charge is large, then it will distort the existing field and the other charges will feel the force change of that. If those other charges are locked in one position, they will still generate the same field contribution as they did before. If those charges are not locked in position, they will move in response to the newly established field, and their new location will result in a changed field.

    What this means is that if you want to use a test charge to measure an existing electric field, you can use a large test charge if the other charges can't move. However, if those other charges can move, then the test charge must be very tiny so as not to move the position of those charges.

    The case of measuring the field from fixed charges using a large test charge is a little non-intuitive because you are measuring the undisturbed field while it is highly disturbed, yet you get the right answer. The key to the understanding is that, although the total field is disturbed, the field at the location of the test charge is the same as it was before you disturbed it. This is equivalent to saying that the test charge can not exert a force on itself.
     
    Last edited: Apr 5, 2009
  8. Apr 5, 2009 #7

    clem

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    The test charge can polarize nearby matter, walls, the lab table, etc., producing an E field that wasn't there before you introduced the test charge,
     
  9. Apr 6, 2009 #8
    Question here is, why should it disrupt?, as I said in the last para -

    And also to the source charge, so it shouldn't matter.

    And so the force on the unit charge, and if we take the unit charge and test it in the same field, it will generate the same force computed...so where's the inaccuracy with a large test charge?

    Oh...ok...got that thanks.

    Very stupid question I must say, but the last doubt still exists.

    Thanks for clearing till now!:approve:
     
  10. Apr 6, 2009 #9

    jtbell

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    Consider for example a conducting object (e.g. a chunk of metal) with a net electric charge on it. The charges arrange themselves in a certain way so the net force on each of them is zero and the charges are stationary (electrostatics).

    This setup produces an electric field which we want to measure at a certain location by putting a test charge there and measuring the electric force on it. But the test charge itself exerts an electric force on the charges in the conductor, causing them to move around and rearrange themselves, and thereby changing the field that they produce!
     
  11. Apr 6, 2009 #10
    That's ok...I got that, thanks, but the problem in the last para still persists; I mean, suppose I propose a 'uninterrupted field theory' the postulates of which -

    -A field cannot be disturbed by other fields

    -Fields can pass through each other, that is at one point in space, 2 fields with different orientations can exist.

    -The consequence of the presence of 1 or more than one field if causes some effect on an entity (for e.g a charged particle), the net effect will be the vector sum of all the effects by individual fields.

    How exactly do you prove this wrong, and so a similar question, how do you prove that the field of a source charge actually gets disturbed in presence of other charges?

    This 'uninterrupted field theory' and the classical theory; by now what I've seen produces the same practical result; but there lies an advantage with the uninterrupted field theory that it's simpler when given a pictorial representation.
     
  12. Apr 6, 2009 #11

    Ben Niehoff

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    Even if you could hold all the source charges in place, you still need the test charge to be small in physical size, regardless of its magnitude. A large test charge will experience tidal forces, and will only measure the average field within a region, rather than the exact field at a point.
     
  13. Apr 6, 2009 #12
    That problem's solved.

    Maybe I should start a new thread for this...ppl don't know what's the current situation.
     
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