- #1
zenterix
- 488
- 72
- Homework Statement
- Why studying inductors, I realized that I have doubts about the electromotive force in a battery.
- Relevant Equations
- Here is what my notes say
If a battery is in a closed circuit, we have an electrostatic field ##\vec{E}## from the positive terminal to the negative terminal of the battery through the circuit outside the battery. This field generates current.
Inside the battery, the electrostatic field lines go from positive to negative terminals. The current inside the battery, however, goes against this electrostatic field. This happens because of some force that transports the charge carriers. We call this force the source force ##\vec{F}_s## and it has origin in chemical reactions.
The work done per unit charge by this source force is called the electromotive force (emf) of the battery
$$\mathcal{E}=\int_-^+\frac{\vec{F}_s}{q}\cdot d\vec{s}=\int_-^+\vec{f}_s\cdot d\vec{s}$$
My question is about the following
My guess is that once the battery is charged to the maximum terminal voltage, then at this point the charge carriers aren't being moved any longer inside the battery and so the net force on such a carrier must be zero.
Is this the explanation?
If this is the case, then it seems
1) The emf is always the same (as long as the battery is capable of performing the same internal chemical reactions at the same rate, since this means that ##\vec{f}_s## remains the same).
2) The terminal voltage goes down as the battery is used and charge is transferred from positive to negative terminal. Assuming that 1) is true then at this point it is no longer true that ##\vec{E}=-\vec{f}_s##. For this to be true we would need to let the battery charge up so that ##\vec{E}## would increase in magnitude to the value of the magnitude of ##\vec{f}_s##.
Is this correct?
Inside the battery, the electrostatic field lines go from positive to negative terminals. The current inside the battery, however, goes against this electrostatic field. This happens because of some force that transports the charge carriers. We call this force the source force ##\vec{F}_s## and it has origin in chemical reactions.
The work done per unit charge by this source force is called the electromotive force (emf) of the battery
$$\mathcal{E}=\int_-^+\frac{\vec{F}_s}{q}\cdot d\vec{s}=\int_-^+\vec{f}_s\cdot d\vec{s}$$
My question is about the following
Why is it that we can say that the sum of the forces is zero?Inside our ideal battery without any internal resistance, the sum of the electrostatic force and the source force on the charge is zero.
$$q\vec{E}+q\vec{F}_s=\vec{0}$$
Therefore, the electrostatic field is equal in magnitude to the source force per unit charge but in opposite direction.
$$\vec{E}=-\vec{f}_s$$
My guess is that once the battery is charged to the maximum terminal voltage, then at this point the charge carriers aren't being moved any longer inside the battery and so the net force on such a carrier must be zero.
Is this the explanation?
If this is the case, then it seems
1) The emf is always the same (as long as the battery is capable of performing the same internal chemical reactions at the same rate, since this means that ##\vec{f}_s## remains the same).
2) The terminal voltage goes down as the battery is used and charge is transferred from positive to negative terminal. Assuming that 1) is true then at this point it is no longer true that ##\vec{E}=-\vec{f}_s##. For this to be true we would need to let the battery charge up so that ##\vec{E}## would increase in magnitude to the value of the magnitude of ##\vec{f}_s##.
Is this correct?
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