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Zero electric field within wire of zero resistance. 
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#1
Oct913, 09:08 PM

P: 47

If a resistor of a resistance = 6 ohms is connected to 2 wires of zero resistance (0 ohms) in a DC circuit with a battery of voltage = 12V, the current throughout is 2A. The voltage drop across a 0 ohm wire is V = R*I. Because R = 0, the voltage drop across the wire is 0V. The only way for this to be possible is for the Electric field inside to be zero. Am I visualizing this right?



#2
Oct913, 09:52 PM

P: 901

Yes.



#3
Oct1113, 05:28 PM

P: 853

excuse me to hijack this thread a little but since the OP got his well thought out question answered I hope you don't mind.
What drives the electrons then if the electric field is zero? And is it then that the higher the resistance of a conductor , the higher the electric field ? 


#4
Oct1113, 05:43 PM

Sci Advisor
Thanks
PF Gold
P: 12,253

Zero electric field within wire of zero resistance.
If there is no resistance then no energy is needed to cause the charge to move. You need to remember that the Kintic Energy of the charges (electrons) does not play a part in this idealised process so no energy is needed to kick the electrons across a wire of no resistance.
When you have a finite resistance then, of course, there is a PD across it. As for the field, that (the Volts per Metre) will depend upon the length and the path of the wire. When considering what goes on in circuits, the fields (in the large scale) are not really relevant and don't usually come into it. 


#5
Oct1113, 11:58 PM

P: 47

They way I think of it is like this:
You initially have two wires connected to the same battery; but they are not connected (incomplete circuit). When the Circuit is connected we assume that the surface charges move in some manner (creates a distribution) which creates zero electric field within the non resistive wires and a large electric field in the resistor. Because there is no field in the wire they are initially stationary in wire. However the electrons in the resistor are not. They experience a force and move in a direction. The velocity of these electrons is constant because the resistance forces act similar to air resistance. When the electrons exit the resistor they encounter the "stationary" electrons in the wires. The electrons exiting the resistor transfer their momentum to the stationary ones. Because all electrons have the same mass, in a 1 dimensional case, all their momentum is lost and is transfered to the stationary ones. Think of this like an elastic collsiona subatomic newton's cradle. In the wire behind the resistor, the electrons in the resistor in front are moving away. This exposes positive charges in the resistor that attract nearby electrons in the wire. The electrons then move into the resistor. This may appear as a force in the wire, but as soon as electrons begin to move in this manner the net force becomes zero when they all move with the same velocity. 


#6
Oct1213, 03:09 AM

P: 853

ok from all that is said , I want to clear up an old misunderstanding on my side.
so zero resistance , zero electric field , now if we have some conductor with some finite value of resistivity then measuring the potential difference at the beginning of the wore and at the end we would get different results , lower voltage the further we go down the length of the conductor , now since we get lower voltage that tells me there also has to be a lower/decreasing electrical field strength the further we go in a conductor with non zero resistance.? Now in a conductor with finite resistance but infinite length we would get zero field and zero volts at the end right? Now then what do we get with an infinitely short conductor with an infinitely large resistance? Since the resistance is infinite we should get also zero volts after it but since it is infinitely small the distance from the " high" side where there is electric field of certain strength and voltage is very small , now how that affects the outcome? Assuming there is no electric discharge or short circuit across the conductor ofcourse. 


#7
Oct1213, 03:58 AM

P: 1,212

What is the definition of voltage. E = grad(V). So a linearly decreasing voltage implies a constant electric field.
I'm struggling to understand what "an infinitely short conductor with infinitely large resistance" could mean. If it has an infinite resistance then it is in no way a conductor. 


#8
Oct1213, 04:57 AM

P: 853

ok, yes right that would be a perfect insulator , okay infinitely small or for practical purpose very short , small conductor with high resistance compared to one which is very long and with the same resistance as measured at both ends.
The question is how do these two differ since the electric field falls of with distance obeying the inverse square law ? The electric field you say is constant all the way to have a linear voltage drop as we see in a resistor, but at the end of the resistor you get a lowered voltage , is that because the field has got weaker due to the drop in voltage ? I guess not , only why not? So if it does imply a constant field then I have a question , put a metal sphere (similar to a point charge) at both ends of the resistor , now as much as I know you should get a stringer charge on the one which is before the resistor than on the one which is after it, yet the field is constant all the way through the resistor , this kinda amuses me. 


#9
Oct1213, 12:44 PM

P: 1,212

Can you describe this comparison precisely? I still don't know what is being compared and for what purpose
Resistance is a quantity you assign to an object equal to V/I. That's all it is. Your objects can be made of silver, aluminium, wood, air, whatever, if the resistances are the same then the same current will flow due to the same potential difference. I have no idea what the last paragraph is, you're not just analysing a resistor in isolation are you? You're presumably connecting it to a circuit, so where do these spheres go? How is a metal sphere similar to a point charge? I think you should find a book on EM and read it from the beginning, these hypothetical questions seem an exhausting way to learn a subject. 


#10
Oct1213, 01:49 PM

P: 47

ok. the (change in voltage)/current = Resistance. Below we examine a circuit constructed of a short conductor of resistance R and long conductor of resistance R. Because they are connected to a battery of the same potential V = 4 the potential difference (E*d) will have to be the same for both the short and long conductors to produce the same potential difference, 4.
so if we have a very long wire with some resistance, the electric field will be small. But b/c there is a long distance E(small)D(long) = change in voltage = 4 Now with a resistor of the same resistance with shorter length. The electric field will be greater to arrive at the same change in voltage: E(large)D(small) = change in voltage = 4. Mathematically an infintely long conductor will need an infinitely small electric field and an infintely short conductor will need an infinitely large electric field...but this is not a physical possibility. 


#11
Oct1213, 03:31 PM

Sci Advisor
P: 2,470

In a conductor with no voltage or thermal gradient applied, the Fermi surface is a sphere centered about zero momentum. So the ground state of the conductor has zero average current. And that means that any net current will quickly die out. This is why an ordinary conductor with zero resistance simply isn't going to happen. On the other hand, we have superconductors. But these are completely different beasts with completely different ground state, and any classic picture needs to be thrown out of the window. To be honest, I'm having very hard time picturing what happens at a junction of a superconductor and a conductor myself. But within the superconductor, the magnetic and electric fields are both zero. 


#12
Oct1213, 04:09 PM

P: 853

another thing I could just add is that because the electric field s the actual source which does the work on those electrons when current flows , we can safely say that long before those real electrons get off the resistor and hit some other electrons in the wire ahead of them the electric field will be already there and will have already done it's job on " pushing" the electrons to where they need to move.
So the bit with the colliding electrons may be a bit off here. 


#13
Oct1213, 08:13 PM

P: 47




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