We know that there is an electric field inside the battery that works against the moving electrons of a circuit. But there is also the chemical force of the battery that at some point become equal. Voltage drop is the integral of the electric field over a closed loop. But you must also have the same integral over the same loop for the chemical force field. EMF is called precisely that, the integral of the chemical force field over a closed loop(the loop with the battery inside). So, can you give me an answer to "Why is the voltage of a battery equal to the EMF?" and having the forces and the integrals in your explanation? Where does the integral of the electric field go? I might be missing or misunderstanding some very basic things about work and voltage, so excuse me if this is the case!
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$\begingroup$ The usual definition of the electromotive force is not as the integral of chemical forces, but as the sum of the integral of the Lorentz force and the effective chemical forces. $\endgroup$– ACuriousMind ♦Mar 20, 2015 at 15:09
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$\begingroup$ I have read that the ΔV of the battery is equal to the integral of the electric field and that is equal to the integral of the chemical forced divided by the charge over a closed loop.I can not understand this.There are chemical forces and coulomb forces.So why isnt ΔV equal to the integral(chemical forces/q + E)dl over a closed loop? $\endgroup$– TheQuantumManMar 20, 2015 at 15:16
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$\begingroup$ $\Delta V$ is restricted in meaning. It's cause is fields set up by charges. It does not include the chemical forces. Those are accounted for by the EMF. $\endgroup$– garypMar 20, 2015 at 15:45
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$\begingroup$ yes,but why the emf is equal to the ΔV?Ι know the thngs that you meantioned $\endgroup$– TheQuantumManMar 20, 2015 at 15:46
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$\begingroup$ @ACuriousMind You get an EMF for every source, a chemical source gives a chemical EMF, a changing magnetic field gives an electric EMF, a moving circuit in a magnetic field give a magnetic EMF. You can have EMFs due to many sources. The EMF due to the chemical source is indeed the line integral of the chemical force per unit charge around the circle, by definition. It's not the total EMF due to everything in the universe, but it is the EMF due to the battery. $\endgroup$– TimaeusMar 21, 2015 at 18:37
3 Answers
Imagine a free-standing battery (not connected to any wires) and take a closed loop through the battery, out one terminal, and back in the other terminal. The total work done in moving a test charge around that loop must vanish. For this to happen, the change in electric potential outside of the battery must equal the negative of the EMF change within the battery. $$\int_\mathrm{outside}\vec{E}\cdot{\mathrm{d} \vec{\ell}} + \mathcal{E} = 0$$
Update after comments
Work is well defined as the integral of force over distance. The relationship between work and energy is more subtle. One needs to carefully define what the system in question is. We also have to recognize that potential energy is the energy associated with the configuration of a system of interacting entities. One is on thin ice if one refers to "the potential energy of a particle". Particles do not have potential energy. The system comprising the particle and something with which it interacts has potential energy. (A ball does not have potential energy. The earth-ball system has potential energy.) I'll review the background on this, with apologies if the background is already well-understood.
Once a system is defined, energy can be added to the system by an external force which can do external work on the system. Work is one way energy can be added. Heat is another, but we'll mostly ignore heat and thermal energy. Generally, $$W_\mathrm{external} = \Delta E$$ where $E$ is the total energy of the system. External work causes energy to be added to the system, but once inside that energy could be potential, kinetic, thermal, chemical ...
Potential energy is defined to be the negative of the work done by forces internal to the system:$$W_\mathrm{internal} = -\Delta PE$$
Now our system. Let's take it to be the wire, the battery terminal, the conductors inside the battery, but not the chemicals and processes that generate the "chemical force". The chemical processes are a source of energy, so we'll take it to be outside of our system. The work done by the chemical processes are external, and do external work on the charge carriers $$W_\mathrm{external}=q\mathcal{E}$$ But the internal voltage due to the separated charges within the battery, also do work, but this work is internal to our system, and thus changes the potential energy of the system $$W_\mathrm{interal} = -\Delta PE = -qV$$ but recall $$W_\mathrm{external} = \Delta E = \Delta PE = -W_\mathrm{internal}$$ (ignoring stores of energy other than potential energy within the system). Finally $$q\mathcal{E}=qV$$ $$\mathcal{E} = V$$
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$\begingroup$ but this means that no charges are moving inside the battery.But if you connect the battery to a wire,then you have movement.Then what can you say about the emf and the integral of E? $\endgroup$ Mar 20, 2015 at 15:47
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$\begingroup$ No, it doesn't. It says that the charge distribution is static. Charges can still move. If a charge carrier leaves the top of the conducting positive electrode, another charge immediately takes its place at the bottom of the positive electrode. $\endgroup$– garypMar 20, 2015 at 15:49
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$\begingroup$ yes,you are right.but how can you explain this with work done by forces? $\endgroup$ Mar 20, 2015 at 15:57
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$\begingroup$ Apart from that,i read in a book that inside the battery the coulomb forces equal the chemical forces.But if this is the case,then how can charges move inside the battery?And if this is the case,how can emf equal the integral of E? $\endgroup$ Mar 20, 2015 at 15:59
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$\begingroup$ The EMF "pumps" the carriers until the electrostatic force is equal to the pumping force, then the net pumping stops. If a circuit removes a carrier from one terminal, thus reducing the electrostatic force, the pump immediately replaces it. Suppose carriers are removed at such a rate that the pump cannot replace them quickly. That means that the number of carriers at the terminal will decrease: the EMF is lower, but the voltage will also be lower. (Remembering that EMF is related to the work needed to move a charge across the terminals, but pumping stops when the forces balance.) $\endgroup$– garypMar 20, 2015 at 16:10
First let's establish the situation in which the result actually holds. Voltage itself is only well defined in electrostatics, and this result only holds in a steady state.
In an ideal battery, there is no energy loss inside the battery during operation, and in the steady state just as much charge flows into the battery as flows out of the battery, and just as much current flows into the battery as flows out of the battery, so the average work done per unit charge inside the battery by both the electrostatic force per unit charge $\vec{E}$ and the chemical (or more generally the source) force per unit charge $\vec{f}_s$ is zero. So if the battery has terminals at a and b then:
$$0=\int_a^b\left(\vec{f}_s+\vec{E}\right)\cdot d\vec{\ell}.$$
Therefore we get
\begin{align}V &=-\int_a^b \vec{E} \cdot d\vec{\ell}\\ &=\int_a^b \vec{f}_s \cdot d\vec{\ell}\\ &=\oint \vec{f}_s \cdot d\vec{\ell}\\ &=\mathscr E_{battery}. \end{align}
The first equality is by an electrostatic definition. The second equality is from our previous equation about the steady state. The third is because the battery only exerts a force per unit charge inside the battery. The last is by general definition of the EMF generated by the battery when $\vec{f}_s$ is the force per unit charge exerted by the battery.
Engineer's (practical) answer: They are the same thing by different names.
EMF and voltage are interchangeable terms as applied to sources.
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$\begingroup$ In most cases, answers with more details tends to better received. Later, after 50 reputation points, you will be able to give comments, too. $\endgroup$– peterhJul 30, 2016 at 9:05