Some theories concerning potential difference

[flyingpig]

Homework Statement

In my book, it says the following

http://img849.imageshack.us/img849/388/unledpx.th.png

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If you go from a to b, you gain potential.

Now here is my question: I thought the E-field points from + to - in that cell, so if the current is going against the E-field, why would you gain potential?

 

[berkeman]

Homework Statement

In my book, it says the following

http://img849.imageshack.us/img849/388/unledpx.th.png

Uploaded with ImageShack.us

If you go from a to b, you gain potential.

Now here is my question: I thought the E-field points from + to - in that cell, so if the current is going against the E-field, why would you gain potential?

Remember that electrons are actually what are being pumped by the battery action against the E field...

Since the E field "points" from + to - in the battery, which way would the electrons prefer to go if they weren't being pumped against that E-field?

 

[flyingpig]

please...

 

[berkeman]

Remember that electrons are actually what are being pumped by the battery action against the E field...

Since the E field "points" from + to - in the battery, which way would the electrons prefer to go if they weren't being pumped against that E-field?

I've undeleted a post that I made a day or two ago. I'm not sure I'm technically accurate in what I've said, but maybe it will help the discussion...

 

[ideasrule]

You do gain potential by moving against the electric force. The electric force wants to push positive charges towards the - terminal, so moving towards the + terminal results in higher potential energy.

Think about a ball on a hill. Pushing it up the hill (against gravity) increases its gravitational potential energy, because it can "potentially" fall back down and gain kinetic energy. Pushing it down the hill decreases its potential energy, for the opposite reason.

 

[flyingpig]

I ask this becuase of this video



Flip to frame 11:27 or so. Mr. Walter Lewin (I really respect this man lol), went from negative to positive and he minused the voltage drop[

 

[gneill]

In that video he states quite categorically that he is not applying KVL. He is applying Faraday's law, and he is summing the integrals of E.dl for each component as he goes around the loop.

 

[flyingpig]

AH shoot this wasn't the only time he did it like this, I can't find you the video right now because I am loaded with finals. Give me some time

 

[SammyS]

$$\Delta V=-\int\vec{E}\cdot d\vec{\ell}$$

Notice the negative sign.

Added in Edit:

The above is in answer to your Original Post, not a response to Lewin's video.

 

[jhae2.718]

Expanding a bit more, in the steady state approximation (i.e. $\partial_t \Phi_b = 0$), Faraday's Law is:
$$\oint\!\! \mathbf{E} \cdot d\mathbf{l} = 0$$
Since $V := -\int\!\!\mathbf{E} \cdot d\mathbf{l}$, this is the same as KVL but with the opposite sign. (In Faraday's Law, the integral is over the closed path of the circuit.) More concisely, if you are familiar with vector calculus, it can be written that $\mathbf{E} = -\nabla V$.

When using Faraday's Law instead of KVL, you can use the other condition of the steady state in place of KLL, namely that:
[tex]\oint\!\!\mathbf{j} \cdot d\mathbf{S} = 0[/itex]<br /> Here, the integral is over a closed surface S, and j is the current density vector. <br /> <br /> When you introduce B-fields into circuits, Faraday's Law becomes necessary, as the electric field in the circuit is no longer conservative. Thus, we have:<br /> $$\oint\!\! \mathbf{E} \cdot d\mathbf{l} = -\frac{\partial \Phi_B}{\partial t}$$<br /> where $\Phi_B$ is magnetic flux, $\Phi_B = \oint\!\!\mathbf{B} \cdot d\mathbf{S}$. (The integral for magnetic flux is over a closed <i>surface</i>...)[/tex]

 

[flyingpig]

Expanding a bit more, in the steady state approximation (i.e. $\partial_t \Phi_b = 0$), Faraday's Law is:
$$\oint\!\! \mathbf{E} \cdot d\mathbf{l} = 0$$
Since $V := -\int\!\!\mathbf{E} \cdot d\mathbf{l}$, this is the same as KVL but with the opposite sign. (In Faraday's Law, the integral is over the closed path of the circuit.) More concisely, if you are familiar with vector calculus, it can be written that $\mathbf{E} = -\nabla V$.

When using Faraday's Law instead of KVL, you can use the other condition of the steady state in place of KLL, namely that:
[tex]\oint\!\!\mathbf{j} \cdot d\mathbf{S} = 0[/itex]<br /> Here, the integral is over a closed surface S, and j is the current density vector. <br /> <br /> When you introduce B-fields into circuits, Faraday's Law becomes necessary, as the electric field in the circuit is no longer conservative. Thus, we have:<br /> $$\oint\!\! \mathbf{E} \cdot d\mathbf{l} = -\frac{\partial \Phi_B}{\partial t}$$<br /> where $\Phi_B$ is magnetic flux, $\Phi_B = \oint\!\!\mathbf{B} \cdot d\mathbf{S}$. (The integral for magnetic flux is over a closed <i>surface</i>...)[/tex]

$$<br /> You are not making this easier...$$

 

[jhae2.718]

Basically, Faraday's Law and KVL are equivalent in certain cases* with the exception that the answers will have different signs.

The difference in signs comes from the definition of electric potential in terms of electric field, which is:
$$V = -\int\!\!\vec{E}\cdot d\vec{l}$$.

*Where there is no time-dependent magnetic field.

 

[jhae2.718]

Also, I forgot to mention that the signs of the circuital elements using Faraday's Law is dependent on the direction you take the path integral. (E.g. the signs will be opposite if you integrate CW or CCW.)