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Charge and potential

  1. May 16, 2015 #1
    Why is q(charge) directly proportional to v(potential)
     
  2. jcsd
  3. May 16, 2015 #2

    Dale

    Staff: Mentor

    If you double the charge then you double the potential. Is there some reason you would you expect any other relationship?
     
  4. May 16, 2015 #3
    Potential is something what created on a point due to a source charge...what is it exactly first...?
     
  5. May 16, 2015 #4

    Dale

    Staff: Mentor

    What do you already know about the definition of potential? What do you think the answer to your own question is, and what makes you uncomfortable with that answer?
     
  6. May 17, 2015 #5
    Potential(kq/r) and is something created on a point due a source charge and if we place a charge at tht point then multiplication of potential wid that chrge make it potential energy
     
  7. May 17, 2015 #6
    What is that something created on the point
     
  8. May 17, 2015 #7

    Dale

    Staff: Mentor

    Yes, the potential tells you how much potential energy a test charge would have if placed there.

    The electromagnetic field.
     
  9. May 17, 2015 #8
    And how is q directly proportional to v
     
  10. May 17, 2015 #9

    Dale

    Staff: Mentor

    I already answered this. If you want a different response then you are going to have to provide some more information that will help me understand your question better.

    What relationship would you expect? What do you feel was insufficient about the previous answer?
     
  11. May 17, 2015 #10

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    There is plenty of experimental evidence to support this.
     
  12. May 17, 2015 #11
    Before that i guess potential differnce confuse me a lot...it is something created in bringing a charge from one point to another point...that means a single charge move say move from point A to B then firstly i would like to ask if it is potential differ at point A or of at B and secondly if potential differnce is created due to change in place of one charge only then why is it said that "potential difference b/w plates"...
     
  13. May 17, 2015 #12

    Dale

    Staff: Mentor

    You can represent the electromagnetic field of a charge in terms of the fields themselves or in terms of the potentials. They are simply two equivalent ways of describing the same thing. As you move charges from A to B you change the fields and therefore you change the potentials also.

    I am not sure what you are asking about regarding the plates. Perhaps you are asking about the charges on the plates of a capacitor and the field between the plates.
     
  14. May 17, 2015 #13
    I take it you are asking about the charge and potential at the same coordinates in space and time.

    In general, charge is not proportional to electric potential. Given a particular electric potential, several terms contribute to charge. Most vanish if the potential is uniform over space or magnetic fields constant in time.
     
    Last edited: May 17, 2015
  15. May 18, 2015 #14
    Yes am asking so
     
  16. May 18, 2015 #15
    Measure the electric potential at some given point, [itex]\phi_1[/itex] due to a bunch of isoloated, static charges, Q1. Replace them with another random bunch Q2 and measure the potential, [itex]\phi_2[/itex]. The resultant potential from the combination Q1+Q2 will be the sum. [itex]\phi=\phi_1+\phi_2[/itex].

    If you double all the charges the potential will double. [Edit: I see Dale has said this, already.]

    This is an example of the superpostion principle for electromagnetism. Mathematically, this worked out because electric charge is just a second derivative of potential in this example. The derivative operator is a linear function.

    The electric charge--more accurately, the charge density is fully described by the divergence of the electric field (Gauss Law). The electric field, in turn, is the difference of the gradient of the electric potential and the divergence of the magnetic potential. All linear functions.
    [tex]\rho=\nabla E[/tex] In terms of potentials, [tex]\rho=\nabla \frac{dA}{dt}-\nabla^2 \phi[/tex]
    give or take a few negative signs. [itex]\rho[/itex] is the charge density and A is the magnetic potential.
     
    Last edited: May 18, 2015
  17. May 26, 2015 #16
    Thank u for giving ur time to answer :)
     
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