Does a constant electric flux produce a current?

[eoghan]

Hi! Suppose I have a wire with a potential difference V=bx where b is a constant. Then there is an electric field which is constant: E=b through out the wire. Well, now consider this relation:
I=\frac{dq}{dt}=\frac{d}{dt}\epsilon_0\int{\vec{E}\cdot\hat{n}dS}=\epsilon_0 S \frac{dE}{dt}=0
where I used the Gauss' theorem and the fact that E is constant.
From this relation then follows that the electric current is 0! But I do have an electric current in the wire! Where am I wrong?

 

[Zaphys]

What surface are you integrating over, what does "x" stand for? Could you explain the problem a little more so that I can help you, personally I can't see it. Thanks

 

[Meir Achuz]

Your Gaussian surface must be closed. The current flows in one side and out the other, which is why the integral is zero.

 

[eoghan]

What surface are you integrating over, what does "x" stand for? Could you explain the problem a little more so that I can help you, personally I can't see it. Thanks

Let's say that the wire lies along the x axis.. then x represents a point of the wire. I integrate over a surface perpendicular to the wire in a point x with the center on the wire. But the surface is meaningless, what matters is that if E is constant in time, then its flux will be constant in time (if the surface is constant in time).

The current flows in one side and out the other, which is why the integral is zero.

Yes, but if you define I=dq/dt then you have I=d/dt (flux(E)) which is zero. With the Gauss' theorem I don't get the net current in the surface (which is zero), but the charge.

 

[Meir Achuz]

Have it your way, but I thought you said "I used the Gauss' theorem".

 

[Born2bwire]

Let's say that the wire lies along the x axis.. then x represents a point of the wire. I integrate over a surface perpendicular to the wire in a point x with the center on the wire. But the surface is meaningless, what matters is that if E is constant in time, then its flux will be constant in time (if the surface is constant in time).


Yes, but if you define I=dq/dt then you have I=d/dt (flux(E)) which is zero. With the Gauss' theorem I don't get the net current in the surface (which is zero), but the charge.

clem is correct here, you used Gauss' Law to relate the charge to the electric field, but Gauss' law requires a closed surface and since you have a spacially constant electric field you will always have zero flux. You should instead try the Lorentz force, however, you would need knowledge of the saturation velocity of charges, otherwise the constant force over an infinite wire will create infinite current.

Still, in respect to your original question, yes, a constant flux can create current. I know that in a simple MOSFET model, we assume that there is a constant electric field between the source and drain which results in a current. The amount of current is limited by the finite cross-sectional area of the transistor, the finite length of the transistor, and the finite saturation velocity of the mobile carriers.