Voltage and electric field in circuits

[proton]

Let's say I have a closed circuit with 1 resistor. Let's say the EMF is between points c and a, the resistor between points a and b, and then between b and c there is nothing but the circuit wire. According to my textbook, the voltage at c is V1, increases by EMF to V2, where V2 = V1+EMF, and then V2 decreases by IR to return to V1 which implies that IR = EMF.

I understand all this. I understand that the voltage must return to V1 after completing the loop. But as you move along the wire where the voltage is a constant V1, deltaV must be 0 which implies that E must be 0 along the circuit path as deltaV is the work done by E. But since a current exists and is the same throughout the circuit, this implies that E must exist. So my question is how can deltaV be zero but the E not be 0?

 

[ranger]

It is because potential difference between the points b and c in this case is the integral of the electric field E:

$$V_b - V_c = \int_b ^c E \cdot d\mathbf{l}$$

What happens when we evaluate an integral whose limits are the same?

 

[proton]

umm... b and c aren't the same points. The whole wire between b and c are at the same potential but the electric field between them is not 0.

 

[ranger]

umm... b and c aren't the same points. The whole wire between b and c are at the same potential but the electric field between them is not 0.

I did not say that b and c were the same points. If the whole wire between b and c is at the same potential, therefore Vb - Vc = 0. This means that Vb = Vc, hence the limits of the integral are the same.

 

[proton]

if the limits are the same, that implies that the integral is 0, but E can still be nonzero as the limits are the same, right? I think I got it. Thanks a lot!

 

[ranger]

Yup, you've got got it. And you're welcome.