11
\$\begingroup\$

When we connect two conductors together in a junction, do electrons actually move from one conductor to another?

Does it also affects the masses of the two conductors?

\$\endgroup\$
7
  • 3
    \$\begingroup\$ Are you talking about the movement of charge once the connection is formed or about the spark that might appear when you make the junction? \$\endgroup\$ Jan 20 at 0:03
  • 1
    \$\begingroup\$ @SredniVashtar I guess both. \$\endgroup\$ Jan 20 at 0:15
  • \$\begingroup\$ physics.stackexchange.com/questions/190591/… \$\endgroup\$
    – Ben
    Jan 20 at 8:50
  • 2
    \$\begingroup\$ @tlfong01 I would like to downvote your answer, but ... oh well. \$\endgroup\$
    – CGCampbell
    Jan 20 at 11:42
  • 1
    \$\begingroup\$ @tlfong01 please don't try to answer questions in the comments. I suggest you delete the comments and post as an answer. \$\endgroup\$ Jan 20 at 15:03
25
\$\begingroup\$

Yes, actual electrons do move through conductors and yes, they cross boundaries like connections between two wires.

While electrons do have mass, there is no net addition of electrons through a conductor. If N electrons are added at one end you will get N out the other end for a net mass change of 0.

Keep in mind that electrons move in a reverse direction (i.e. from - to +) which is the opposite of the conventions we normally use for current from which is from + to -. Also note that electrons move quite slowly through conductors even though the electric current moves much faster. The electrons themselves move at something like 1 mm/sec in a copper wire.

\$\endgroup\$
8
  • 2
    \$\begingroup\$ there is no net addition of electrons through a conductor - In the electrostatic case, there is. \$\endgroup\$
    – Reinderien
    Jan 19 at 23:33
  • 2
    \$\begingroup\$ Yes, that's true but I didn't get the impression that we were talking electrostatics here. \$\endgroup\$
    – jwh20
    Jan 19 at 23:39
  • 1
    \$\begingroup\$ Let's assume you have two objects, one has an electrostatic charge. By that we mean that electrons have been "forced" onto the object by some charge generator. Perhaps a Vandegraff Generator or just some carpet. We have another object with no net charge. If you touch them together the electrostatic potential will equalize and electrons will flow from the more highly charged one to the other one. Yes, the mass of each object will change in proportion to the amount of electrons that moved. \$\endgroup\$
    – jwh20
    Jan 20 at 0:02
  • 4
    \$\begingroup\$ Bear in mind an electron's mass is about 1/100000000 of, say, the copper atom that loses it so you're really not going to notice the change. \$\endgroup\$
    – Finbarr
    Jan 20 at 0:08
  • 2
    \$\begingroup\$ @VolkerSiegel that's why it's called drift velocity. \$\endgroup\$
    – Arsenal
    Jan 20 at 9:37
11
\$\begingroup\$

Just a short answer to get you thinking ...

When we connect two conductors together in a junction, do electrons actually move from one conductor to another?

Electrical current is the flow of charge. This is stated neatly in the equation $$ I = \frac {dQ}{dt} $$

where I is current, Q is charge and t is time.

Now consider a DC circuit with a switch. When you close the switch the current will be the same on both sides of the switch and in the same direction. It stands to reason then that the flow of charge must cross the contacts. If the resistance of the contacts is high - as is often the case due to the small contact area - then the contacts get hot as we would predict from the power equation $$ P = I^2 R $$ where R is the contact resistance.

Does it also affects the masses of the two conductors?

No. No additional mobile charge carriers are introduced. We just push the existing charge carriers along. A (poor) analogy is a bicycle chain: it transmits power but the number of links doesn't change and it's weight doesn't change. To compliment @jwh20's point about the speed of the electric wave relative to the speed of the charges, the effect of standing on the pedals is felt immediately at the back wheel even though it takes a second or two for the links to travel from the rear sprocket to the front chain ring.

\$\endgroup\$
8
  • \$\begingroup\$ Thanks a lot for the explanation! What if you have two charged objects and we touch them together - will it change the masses of the objects when electrons move from one of them to the other? There is no closed loop in this case. \$\endgroup\$ Jan 19 at 23:46
  • \$\begingroup\$ Another follow-up question about current going through I high resistance contacts of a switch. Shouldn't current decrease in this case (since resistance is too high but voltage didn't change?). \$\endgroup\$ Jan 19 at 23:48
  • \$\begingroup\$ @RodionDegtyar No, it's the current that stays the same. There is a voltage drop across the contact resistance in accordance with Ohm's law. \$\endgroup\$
    – Finbarr
    Jan 19 at 23:56
  • \$\begingroup\$ "What if you have two charged objects and we touch them together ..." That's a different question so I won't answer it here. Ask a new question after you've done your research. \$\endgroup\$
    – Transistor
    Jan 20 at 0:00
  • \$\begingroup\$ This is a short answer? How is my contact Resistance defined if not by pressure et temperature? Electrical current is the flow of charge Amerage is what hurtz not voltage or resistance \$\endgroup\$
    – Jay
    Jan 20 at 7:36
5
\$\begingroup\$

Electrons have an appoximate mass of \$10^{-30}\text{ kg}\$.

Even so, when current flows in a loop, the electrons bump in one end and bump another out the other end of the conductor, so no change in mass occurs.

Electrons can jump across contacts when the voltage is different when contact is made.

When the current is high enough you can hear it and maybe see it. Such as plugging in a mobile charger. Since it is a current loop , the jump occurs on the last of 2 contacts made which is normally V+ by design.

This sound of the arc current is actually occurring so fast that it breaks the sound barrier and makes a tiny tick or a Big Bang such as thunder from lightning. This is simply from capacitance charge or discharge and the negative resistance from ionization of the arc between the conductors.

\$\endgroup\$
5
  • \$\begingroup\$ It will not break the sound barrier under vaccum and in a other such manifold such force would result in ultra luminescence and therefor produce light not sound or thunder \$\endgroup\$
    – Jay
    Jan 20 at 7:41
  • \$\begingroup\$ You don't get an arc in a vacuum as there's nothing to ionise. \$\endgroup\$
    – Finbarr
    Jan 20 at 10:42
  • 1
    \$\begingroup\$ @Finbarr to be exact in vacuum, there will be no spark, but an arc (formed by ions emitted from the electrodes) can exist under some circumstances \$\endgroup\$
    – dominecf
    Jan 20 at 11:00
  • 1
    \$\begingroup\$ @Tony "breaking the sound barrier" is not exact as no sound source actually moves (like a Concorde); you can get a cracking sound just by a sub-sonic rapid expansion (like popping a balloon). \$\endgroup\$
    – dominecf
    Jan 20 at 11:03
  • \$\begingroup\$ If you understand the ionization process and slew rate of the arc in air, you will understand. A balloon is different and much slower pulse \$\endgroup\$ Jan 20 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.