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Re: [Phys-L] [**External**] Re: [**External**] Photoelectric effect: electron current vs. frequency

• From: John Denker <jsd@av8n.com>
• Date: Wed, 10 Nov 2021 11:37:47 -0700

On 11/10/21 11:14 AM, Philip Keller via Phys-l wrote:

"Let's suppose, for simplicity, that electrons are popping out
at the same rate, i.e. the same number of electrons per second.
That is the *definition* of current: charge per unit time.
The velocity has got nothing to do with it."

That's the definition of current out of the target. But is it what the
ammeter in the simulation measures? I thought it measured the rate
electrons reach that second electrode.

I would think current out of the target equals current into
(i.e. through) the ammeter (assuming more-or-less steady
state conditions). Conservation of electrons. More precisely,
and more to the point: conservation of charge.

It is straightforward to rig up the ammeter so that no
electrons go astray.

In particular, the speed with which the electrons impinge on
the ammeter does not affect the ammeter reading.

The ammeter measures the number of electrons per second. What
else could it possibly do? I suspect I don't understand the
question.

In particular, here's a way of building an ammeter that is
easy to understand, although suboptimal in other ways: Let
the current flow into a capacitor. The integral dt of the
current is the charge on the capacitor. So you can measure
the voltage on the capacitor, multiply by the capacitance
to get the charge, and plot charge versus time. The slope
of plot is the current, as a function of time. How hard the
electron smacks into the capacitor plate has got nothing to
do with it.