How fast are electrons moving when they leave the electric wall socket in your house (240V in the UK)?

Each individual electron is moving quite slowly. There's something called the drift velocity which is:

 

[math]v_d = \frac {I} {nqA}[/math]

 

Where I is the current, n is the density of charge carriers, q is teh charge of electrons (charge carrier) and A is the area of the wire.

Did you know, as a result of Dirac co-formulating quantum mechanics and relativity together, he found that the electron could move at light speed along jagged paths through spacetime? The jagged path was in fact caused by it literally ''bouncing'' off negative electrons in the Dirac Sea?

 

Its a beautiful theory.

Did you know, as a result of Dirac co-formulating quantum mechanics and relativity together, he found that the electron could move at light speed along jagged paths through spacetime? The jagged path was in fact caused by it literally ''bouncing'' off negative electrons in the Dirac Sea?

 

Its a beautiful theory.

 

And the relevance to the OP is . . . what?

Thanks for the replies. Why is it ten if electrons are moving slowly that if I use fairly lightweight wire to connect a 2.5kW kettlethen the wire itself gets very warm if not hot and possible melts but if I use a thicker wire all remains cool?

 

It's as if the kettle is sucking the electrons out of the wall socket at a terrific rate compared to say a 100W table light.

That would be because the current used by the kettle is far larger than the current used by the light.

So is it corerct to say that the higher the current of an appliance then the greater the amount of "suck" it has?

I don't think that relates to the speed of the electrons, only their quantity.

So a 2.5kW kettle sucks more electrons out of a wall socket compared to a 100W table light. I ask these questions because I'm just wondering how you get an electric shock. If I stick my finger in the live terminal of my wall socket I'll get a hell of a belt but is my finger sucking any electrons out of the socket?

The current tells you how many electrons are flowing, and the potential (voltage) tells you how much energy they have and will lose in going through the circuit; the power dissipated is the product of the two. For each Volt, you'll dissipate 1 Joule of energy for each Coulomb of charge , or 1 Watt per Amp.

 

A 2.5 kW appliance will draw more current than a 100 W light bulb, since the potential drop is the same (240 V is your case).

As I'm not an electric kettle or table lamp how do I end up getting an electric shock if I put my finger in the live outlet of my wall socket?

There's still a potential difference between you and the socket, it's a constant voltage power supply so it tries to push as much current as possible to make the potential difference 240V....

Why do you not now explain how the displacement "flow" of electrons "pass through" a capacitor, if the source be a.c.?

 

Also, since the dielectric blocking the "flow" of electrons has finite (though very high) resistance, figure on some of the displacement current actually passing through the dielectric, even as excess and deficiency of electrons pile up on either side of it?

 

Finally, why cannot electrolytic capacitors satisfactorily handle a.c. circuits?

 

imp