Electric Field Propagation

  • Thread starter Emanresu
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Hi, I'm trying to get my head around electric field propagation in an electric
circuit. If you have a battery with no circuit I imagine the electric field
between the terminals would look like the magnetic field lines between the
two poles of a magnet. What I can't visualize is what the field lines look
like when you add an open circuit and slowly close it. What does the field
look like when it is nearly closed and when it is fully closed ? Does it only
propagate when the circuit is fully closed ?
 

pervect

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I'm not positive I'm interpreting your question correctly, but if you consider a pair of contacts close to each other, the potential at one contact will be +v, and the potential of the other contact will be ground (zero).

Note that all points on a conductor have the same potential as long as no current is flowing. (If current is flowing, I*R losses in the conductor will slightly affect the potential at different points on the conductor).

The field between the contacts will be v/d, the foltage/distance. Think of the field between the two contacts as the field of a parallel plate capacitor for a rough guide. (You can look that up on the internet if you don't already know what it looks like, or ask another quesiton).

As the distance between the contacts is slowly decreased, the electric field will increase.

When the field gets high enough, the air ionizes, and a spark jumps the gap, though with low voltage batteries (12 volts) this doesn't happen unless the spacing is very, very, very close.
 
53
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I really don't understand fields so please bear with me. You say the field strength is
v/d, so if I attach insulated wires to the battery terminals and hold the ends of the
wires apart by the same distance that was between the terminals do I get a field
of the same strength as was between the battery terminals ? Is surface area relevant ? If voltage determines the field why is it said that it takes time for the field to propagate when a circuit switch closes if the 2 sides of the switch are already at
the battery terminal potentials ?
 
1,675
3
Emanresu said:
I really don't understand fields so please bear with me. You say the field strength is
v/d, so if I attach insulated wires to the battery terminals and hold the ends of the
wires apart by the same distance that was between the terminals do I get a field
of the same strength as was between the battery terminals ?
No the field will be bigger. The electric field is the voltage divided by the
distance between the electrodes.

Is surface area relevant ?
No, but the shape will affect the field. E=V/d is exact for two infinitely
large flat parallel plates seperated by d.

If voltage determines the field why is it said that it takes time for the field to propagate when a circuit switch closes if the 2 sides of the switch are already at
the battery terminal potentials ?
The change in space of the voltage determines the electric field.
It takes time for a disruption of the field to propagate.
 

pervect

Staff Emeritus
Science Advisor
Insights Author
9,554
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Emanresu said:
I really don't understand fields so please bear with me. You say the field strength is
v/d, so if I attach insulated wires to the battery terminals and hold the ends of the
wires apart by the same distance that was between the terminals do I get a field
of the same strength as was between the battery terminals ?
Yes, if I'm visualzing this right, you have the same voltage over the same distance, and thus the same field.
Is surface area relevant ?
The "V/d" calculation is an approximation, assuming fairly large and flat electrodes. Surface area matters to the "fringing fields". If you have very small electrodes, surface area and surface curvature can become very important. (Sharp points are associatied with high fields near the sharp point - this can be seen by using a model based on the electric field from a sphere of radius r and charge q, and varying the radius).

I'd advise you to read up some more a bit on parallel plate capacitors and the electric field from a physics book - a really exact descritpion in a short post is hard to do. Look at the fields due to a sphere, and the fields due to a parallel plate capacitor.

If voltage determines the field why is it said that it takes time for the field to propagate when a circuit switch closes if the 2 sides of the switch are already at
the battery terminal potentials ?
The two sides of the switch are not at the same potential before the switch closes. To describe this as simply as possible - imagine putting a voltmeter across the open switch - or actually do the experiment. You'll see that it reads the voltage of the battery. (This is assuming that there is only one switch in the circuit , I'd draw a diagram but it's too hard to do in ascii).
 
53
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Antiphon said:
The change in space of the voltage determines the electric field.
It takes time for a disruption of the field to propagate.
What is the disruption ? I read somewhere that if you had a circuit with
a battery, bulb and two wires 300000km long and connected the circuit
it would take a second for the bulb to light. Say you had a switch beside
one of the battery terminals that was open by 1mm, then there exists an
electric field in that 1mm gap that wants to move electrons, but nothing
in the rest of the circuit. Now you close the circuit and a second later the
electrons move in the bulb and it lights. What was happening during that
second ?
 
1,675
3
During that second, the change in the electric and magnetic fields are zipping
down the wire toward the bulb. The change started at the switch and moved
from there. Before you closed the switch, the current was zero. But now it's
not zero. It takes time for this "wave front" to reach the bulb.

(For the type of wire you describe, it would take longer than one second because
"signals travel less than the speed of light down a two-conductor transmission line.")
 
53
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Thanks guys, I think I'm about half way there now but I'll have to do some thinking
(and reading) before I come back with my next lot of questions.
 

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