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f95toli

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That formula works for DC as well. An "exotic" example would be a single electron pump where the (dc) current is given by the number of electrons pumped per second...

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I don't believe it does. Say that a constant 10 Coulombs of charge per second is flowing through a conductor. This would be equivalent of 10 Amps of current. The rate of change in Coulombs per second is zero. So the equation i(t)=dq(t)/dt would yield zero also.

That formula works for DC as well. An "exotic" example would be a single electron pump where the (dc) current is given by the number of electrons pumped per second...

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Andrew Mason

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!? The rate at which charges move is dq/dt. That is the current. You appear to be confusing dq/dt with dI don't believe it does. Say that a constant 10 Coulombs of charge per second is flowing through a conductor. This would be equivalent of 10 Amps of current. The rate of change in Coulombs per second is zero. So the equation i(t)=dq(t)/dt would yield zero also.

AM

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No, itThis would be equivalent of 10 Amps of current.

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Say at t = 2 seconds there is 10 Coulombs flowing in the conductor, at t = 3 seconds there is 10 Coulombs flowing in the conductor. A simplified way of figuring dq/dt would be (10 - 10) / (3 - 2) = 0. A slope of a horizontal line is always zero.!? The rate at which charges move is dq/dt. That is the current. You appear to be confusing dq/dt with d^{2}q/dt^{2}. The rate of change of charge flow is not the same as the rate of charge flow.

AM

d

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- #7

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There is or there are?Say at t = 2 seconds there is 10 Coulombs flowing in the conductor

But no, there are 10 coulombs per second flowing.

If you re-examine your units you will understand what everyone is telling you.

- #8

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It helps if you think in terms of current DENSITY at first:I don't believe it does. Say that a constant 10 Coulombs of charge per second is flowing through a conductor. This would be equivalent of 10 Amps of current. The rate of change in Coulombs per second is zero. So the equation i(t)=dq(t)/dt would yield zero also.

Now consider a cross-section of your conductor and perform a surface integral, the result:

I = (dq/dx) (dx/dt) = dq/dt

So the proper interpretation of the dq/dt term is the rate of charges

- #9

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OK I get it now. I was thinking in terms of current flow, not charge flow. At t=2, 20 coulombs have flown through a conductor. At t=3 30 coulombs have flown through a conductor. So (30 - 20) / (3 - 2) = 10 coulombs per second. The coulomb count is accumulative.There is or there are?

But no, there are 10 coulombs per second flowing.

If you re-examine your units you will understand what everyone is telling you.

Sorry for the confusion (and bad grammar) - watching a two-year old can cause a lack of concentration.

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