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In the attachment, which comes from a text "Introduction to Electrodynamics" by Griffiths. There are two statements: "Within an ideal source of emf (resistenceless battery), the net force on the charges is zero, so $\vec{E} = -\vec{f}_s$" and further down at the bottom we have the statement "inside the battery $\vec{f}_s$ drives current in the direction opposite to $\vec{E}$". Would the correct interpretation of this be that the charge has no net force when it moves through the battery (since $\vec{E} = -\vec{f}_s$) and so essentially the charge is moving at a constant velocity within the battery, it entered the battery at an initial velocity from being driven by the electric field force from the electric field in the rest of the circuit (where in the circuit itself it has a net force $\vec{E}$ per unit charge)? Is this the correct interpretation? Or close?

Thanks.

enter image description here

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Electrons do not move through batteries. Not in reality, and neither in this model, I assume.

in 7.9, an integral is taken over the whole (closed) loop. inside the ideal battery, $E=-f_s$. This is not the case in the entire external circuit (there you can only skip the contribution of $E$ because it is an integral over a closed loop). Consider $\int_a^b\mathbf{f}\cdot d\mathbf{l} = 0$ inside the battery. If you extend the integral to the whole loop, you get 7.9. So the emf is really between the two terminals over the external circuit.

The reason an ideal battery is mentioned is because in a real battery $|f_s|>|E|_{ab}$, because of the internal resistance.

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  • $\begingroup$ In this model it does assume that the electrons move through the battery, see highlighted sentence at the bottom which states "notice, however that in the battery $\vec{f}_s$ drives the current in the direction opposite to $\vec{E}$". $\endgroup$
    – Alex
    Dec 7 '16 at 14:00
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To clarify this is the paper Griffiths was citing. It depicts Cd++ and Hd+ flowing towards positive terminals which is where “current is being driven in an opposite direction to E.” enter image description here

I thought the original person who asked this question had a pretty good understanding except that it’s unnecessary to describe current as a single particle traveling through the entire circuit.

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  • $\begingroup$ Where can I find the complete abstract? $\endgroup$
    – user248666
    Oct 4 '20 at 11:00

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