# Emf must be greater than it is supposed to be

Suppose a battery of negligible internal resistance which creates a potential difference V across its terminals is connected to a resistor R through wires of negligible resistance then it creates voltage V across the resistor too.

Its emf is the energy given by itself to the unit charge to move it from negative terminal to positive terminal (current being conventional). Then there are two places where energy is being lost by the charges 1) due to resistance of resistor and 2) due to the work done by electrostatic force due to the electric field between Terminals of battery which opposes the motion of charge as it moves from negative to positive terminal and hence this work is negative.

So,

$$emf × q = (V × q) + (qEd)$$ $$\implies emf= V+Ed$$ Where,

q is the charge on which battery does work in a given time and in the same time energy is being given out by q passing through resistor.

d is the straight distance between two terminals. I have supposed that the charge q moves from negative to positive terminal along the shortest distance between them.

E is the electric field intensity between terminals directed from positive to negative terminal. I have supposed it is uniform at any point between two terminals.

The problem is that my book says different as compared to what I wrote in the equation. It says:

$$emf × q = V × q$$ or$$emf=V$$

The charge doesn't lose energy when it moves inside the two plates. Consider two plates of the battery - positive and negative. The charge moves from the negative plate to the positive plate because of the battery force (just a name for the mechanism due to which the battery works - chemical , mechanical etc). This force is equal to the electric field and the charge experiences no net force between the plates. The battery force does work and there is increase in the potential energy of the charge which is then lost in the resistor. Then $$EMF × q = V × q$$ The battery mechanism provides the energy to be lost in the circuit.