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Question about charge in a circuit

  1. Feb 4, 2009 #1
    I'm unsure about the equation Q = It (charge = current * time). Say we have a perfect circuit with minimal resistance. The current will stay constant and never run out. Work doesn't need to be done because no resistance needs to be overcome. But then, as time goes on, the charge increases, according to Q = It. But how can we keep getting more charge without doing work? What about conservation of charge and all that?
  2. jcsd
  3. Feb 4, 2009 #2
    An amp is one coulomb per second. Multiplying amps by seconds will tell you how many coulombs have passed a point on the conductor. The charge doesn't increase any more than the amount of water increases in a closed circulating system.
  4. Feb 4, 2009 #3
    Yeah… IF

    6.241506×10^18 electrons per second = 1 amp
    1 coulomb = 6.241506 x 10^18 electrons

    Shouldint the equation be

    Charge = 6.241506×10^18 ( Amps * seconds )

    Like amps times seconds times 6.241506×10^18 equals the number of electrons.
    Iv noticed textbooks get things wrong alot so you just have to think about it.
    Last edited: Feb 5, 2009
  5. Feb 4, 2009 #4
    That's right.
  6. Feb 10, 2009 #5
    You're right about everything but this part. As you went on to correctly identify, 6.241506×10^18 ( Amps * seconds ) gives you the # of electrons, but charge isn't measured in electrons, it's measured in coulombs. The conversion isn't necessary.

    As It was said before, an amp is a coulomb per second, therefore n amps is n coulombs per second, hence the expression I=Q/T

    Algebra tells us, then, that Q=IT is true.
  7. Feb 10, 2009 #6


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  8. Feb 10, 2009 #7


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    Think of Q as being the amount of charge that flows past a fixed point in the circuit during time t. The total amount of charge that is moving around the circuit doesn't change. It simply gets pushed around the circuit.

    If the current comes from a battery, then during a given amount of time, as many electrons enter the + terminal of a battery as leave the - terminal, so the total number of moving electrons remains constant.

    If the current is being produced by electromagnetic induction (as in a generator), there is no battery, and we literally have the same electrons going round and round for as long as the induction continues.
  9. Feb 10, 2009 #8
    If charge is measured in coulombs then…

    Charge is measured in electrons, its just units of electrons called coulombs. Following your analogy is just like rewriting my equation as…

    Charge = 1 coulomb ( amps * seconds )

    and that’s the same equation as

    Charge = 6.241506×10^18 ( Amps * seconds )

    Just like every basic text book on the planet will tell you Q = It. This is only true when relating the equation to algebra. Its false when relating the equation to reality.
    Last edited: Feb 10, 2009
  10. Feb 10, 2009 #9


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  11. Feb 10, 2009 #10


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    so I think your claim is

    C = n * e
    C -> charge
    n -> number of electrons
    e -> charge per electron

    but then what n will solve this equation?

    -C = n * e

    clearly you need a second particle. but this second particle can cancel the electron. so, then of the electrons passing by a point per unit of time how many were canceled and how many were not? i think this dilemma will illustrate why I = dQ/Dt != d(number of electrons)/dt

    edit: i guess i should say in the general case. one could always constrain the problem to only electrons as charge carries and then i guess it would be true.
  12. Feb 10, 2009 #11
    AKA, Charge = Amps * Seconds

    An amp is a coulomb per second. So, amps * seconds = coulombs.

    You say Q = coulombs * amps * seconds
    The units don't work out. You would have charge measured in coulombs^2, because of your extra coulombs term.
  13. Feb 11, 2009 #12

    I said Q = amps * seconds * coulumbs
  14. Feb 11, 2009 #13
    Haha, please tell me you know that's the same thing.
  15. Feb 11, 2009 #14
    Haha I ment to say

    Q = amps * seconds * 1 coulumb
  16. Feb 11, 2009 #15
    I see what you're doing. The "1 coulomb" term is a conversion factor.

    That would be fine, but "Q = amps * seconds" is already in terms of coulombs. If you add in another "1 coulomb" term, then Q would be in terms of coulombs^2.

    Also, keep in mind that a coulomb is not a number of electrons. Sure, there is a number of electrons that add up to have -1 coulombs of charge, but "coulomb" isn't a counting unit, like the "mole" is when dealing with molecules. Of course, you can incorporate a "coulombs to electrons" conversion factor to find out how many electrons are flowing through the system. You would use amps * seconds to find Q, which is a number of coulombs, then multiply Q by the term "6.241506 x 10^18 electrons / 1 coulomb" to put it in terms of the # of electrons.

    Just remember that Q is always measured in coulombs, just like V is measured in meters/second. You can use Q to find the # of electrons, but that doesn't mean you're measuring charge in terms of electrons. That would be like measuring length in terms of pencils. Do you know what I'm saying?
  17. Feb 11, 2009 #16
    Yeah I see what your saying and it makes good sense if a coulomb “isn’t a counting unit”.

    But if a coulomb “isn’t a counting unit” equal to 6.241506 x 10^18 electrons then whats your definition of coulomb???
  18. Feb 11, 2009 #17
    You can also think of it this way.

    Consider a circuit. Now take a small section of the wire in that circuit. Call the end point A and B.

    As the current flows through the circuit, how does the charge in this segment change? Easy. Take the current flowing INTO the segmen through A (Q_A) and the current flowing OUT of the sement through B (Q_B) and subtract them. The tricky part is getting the signs correct. If the segment of wire does not contain any capacitors or other oddities, then Q_A = Q_B, so dQ/dt = Q_A - Q_B = 0.

    Really, it's just like level water in a container is equal to the water coming in minus water coming out. Only instead of liters of water, it's coulombs of charge.
  19. Feb 11, 2009 #18
    A Coulomb is the unit we use to measure charge.
    Just like we use Kelvin for temperature, and meters for distance.

    We use coulombs to measure the charge of anything, not just electrons.

    For example, a proton has the equal but opposite charge as an electron.

    An proton has a charge of 1.602176487×10^-19 coulombs.
    An electron has a charge of negative 1.602176487×10^-19 coulombs.

    If coulombs were just a counting unit for electrons, that would mean that a proton was made of negative 1 electrons.

    See? A coulomb is just how we measure charge, which exists in more than just electrons.

    Here is dictionary.com's definition:

    the SI unit of quantity of electricity, equal to the quantity of charge transferred in one second across a conductor in which there is a constant current of one ampere

    Good questions!
  20. Feb 11, 2009 #19
    The coulomb unit works equally well if we treat electronic charge as a quantized or a continuous quantity. I don't know the historical development, but it seems the coulomb was probably defined prior to the discovery of the electron to begin with.
  21. Feb 11, 2009 #20
    The Coulomb was defined in the 1780's by Charles Augustin de Coulomb. This was roughly 100 years before the discovery of the electron, so no, the coulomb isn't defined in terms of electrons.

    This is how it is defined.

    The equation for the gravitational force between two particles and the equation for the electromagnetic force between two particles are very similar. I'll do some color coding to make it easier to understand what I'm saying.

    They both say that:

    (The attractive force) = (A special constant) X (A particular property of the first particle) x (a particular property of the second particle) / (the square of the distance between the particles)

    The particular property of the particles that the gravitational force depends on is mass.
    And The particular property of the particles that the electromagnetic force depends on is charge.

    The "special" constants, for lack of a less adorable word, have to do with the relative strengths of the two forces. Gravity is naturally significantly weaker than the electromagnetic force. (That's why even though a penny can be pulled by gravity all the way down to the street from the top of a building, the repulsive electro magnetic force from the particles in the street can stop it almost instantly) Since the electromagnetic force is so much stronger than gravity, it's constant is larger than gravity's is.

    Here's the important part. This is how the Coulomb was defined:

    If two particles have a charge of the order of one unit charge ( 1 coulomb) and mass of the order of a unit mass (1 kg), the electrostatic forces will be so much larger than the gravitational forces that the gravitational force can be ignored. This works with particles that are any distance apart.

    We don't measure charge in electrons, charge is a property of an electron. An electron isn't in and of it self charge.

    Measuring charge in electrons would be like measuring mass in apples.

    Did this help at all? I realize the way I wrote out the equation may be a little confusing. I promise I'll learn to use that fancy code....
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