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why work done by a battery in circuit is potential diffrence across plates times charge flown through body.

W=Qε(ε is emf of battery)

although there are heat and other loses ? and another that my text book has not mentioned from which point till which point in circuit Q is being transffered?

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Batteries use a chemical reaction to do work on charge in the battery to generate a voltage across its terminals giving the charge potential energy. If the battery is not connected to a circuit that voltage is called an emf which means the voltage across its terminals when there is no current flow (the open circuit voltage).

All real batteries have internal resistance. Once connected to a circuit and current flows there is a voltage drop across its internal resistance. That reduces the terminal voltage and results in some heat loss internally.

Now the battery does work to deliver charge to the circuit. The charge loses the potential energy the battery gave it and it’s either dissipated or stored in the circuit elements. The work it does now is your equation except you replace emf with the terminal voltage because that’s the voltage actually applied to the circuit. While it delivers energy to the circuit it also dissipates some energy internally.

Finally the battery then does work on the charge returning to one terminal once again giving it potential energy at the other terminal.

Hope this helps

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The key here is that the potential comes from the fact that there are losses. Consider a circuit element like a resistor; If instead this was an ideal wire, then the work needed to move a charge (neglecting external fields such as gravity) is 0 because there is no force applied to the charge. Now if we replace this ideal wire with a resistor, there is a force needed to move the charge across the element due to the dissipation of energy (to heat in this case). This is much like pushing a block up a frictionless hill vs across a frictionless table.

We can then derive the expression for work from the definition:

$$ W = \int_a^b \mathbf{F} \cdot \mathbf{\mathop{dr}} $$

Where $\mathbf{\mathop{dr}}$ is the displacement along the path.

Then since $\mathbf{F} = Q \mathbf{E}$ we get:

$$ W = Q \int_a^b \mathbf{E} \cdot \mathbf{\mathop{dr}} $$

But for a circuit element, the integral is the potential difference between $a$ and $b$. So we can write this as:

$$W = QV$

Or, for a battery, where we call the potential difference the EMF $\epsilon$:

$$ W = Q\epsilon$$

As you wanted to show.

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