# Why is source force + electrostatic force = 0 inside battery?

• zenterix
zenterix
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

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.

$$\vec{E}=-\vec{f}_s$$
Why is it that we can say that the sum of the forces is zero?

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?

Last edited:
zenterix said:
Why is it that we can say that the sum of the forces is zero?
Because the work which source force per unit charge does is
$$\int_{minus \ plate}^{plus \ plate} f_s dx =\int_{minus \ plate}^{plus \ plate} - E dx$$
which is positive and minimun among other conditions because it is ideal, without friction, resistance, heat gereration and all the other types of energy dissipation.

Last edited:

(a) ##\vec{f}_s=-\vec{E}## inside the battery.

You say this is because

(b) ##\int_-^+ \vec{f}_s\cdot d\vec{s}=-\int_-^+\vec{E}\cdot d\vec{s}##.

But I don't see (b) as a justification.

Of course if I assume (b) is true then (a) is true but this seems to be the result of integration rules, not physics.

You say that (b) is "positive and minimum among other conditions". Not sure what you mean by this. Sure, the integrals are positive since we assume that ##\vec{f}_s## is a force "pushing" charge carriers along the direction of the trajectory between the terminals.

What do you mean by minimum?

zenterix said:
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.
.
.
2) The terminal voltage goes down as the battery is used
I’m not sure that ‘source force’ is a widely used concept for cells/batteries. Maybe I’ve led a sheltered life. But here’s my ‘physical’ interpretation (assuming a simple DC circuit).

For an ideal battery, terminal-voltage = emf and this value never changes.

That’s because an ideal battery has zero internal resistance and an infinite supply of (chemical) energy. The ideal battery can deliver any size current and respond instantaneously to any changes in the external circuit (with no change in terminal-voltage).

Inside the ideal battery, at any instant, the charges can be considered as flowing in a steady-state, i.e. not accelerating. That means the net force on a charge must be zero. This net force is the vector sum of the electrostatic force and the ‘source force’.

The speed of the charge carriers can be between zero and infinity, depending on the power being delivered to the external circuit. But the electrostatic force and the ‘source force’ have fixed equal magnitudes and act in opposite directions.

[Minor edit.]

PeroK and berkeman
zenterix said:
What do you mean by minimum?
$$f_s=-E+a$$
Charge undertakes acceleration. When charge reach plus plate, it has got surplus kinetic energy of
$$\int a dx$$
Charge collide with plate and dissipate its surplus kinetic energy which would finally be transformed to heat energy. That does not seem to be ideal battery. a=0 is the case where minimum required energy is supplied to charge for going up to plus plate.

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cianfa72
From one point view if we see then in an ideal circuit there is no dissipation of energy and the amount of energy provided by the system i.e. the battery should be equal to the amount of energy gained by the battery , which results no net amount of change in energy resulting zero net force as force is the gradient of energy.

From another point of view we can say that according to Ohm's Law
J = σf where f is the total force including source force and electrostatic force.
In an ideal circuit ρ = 0 ⇒ σ = ∞
f = J/σ = J/∞ = 0

berkeman
What is "J" here?

nasu said:
What is "J" here?
Current Density

Then your formula is wrong. It should be electric field and not force on the right hand side.
##j= \sigma E##

nasu said:
Then your formula is wrong. It should be electric field and not force on the right hand side.
##j= \sigma E##

This is an extract from the book An Introduction to Electrodynamics by David J Griffiths.

Debashis Saikia said:
View attachment 343509View attachment 343510

This is an extract from the book An Introduction to Electrodynamics by David J Griffiths.
Note that ##\mathbf f## in that equation is force per unit charge. I.e. the electric field.

nasu

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