# Reference point of potential in case of a battery

When we define electric potential of a charge,we need to take a charged body as a reference and then we bring the test charge to calculate the electric potential. Now,electric potential energy is actually a function of position with respect to the reference body. And electric potential is simply the electrical energy per unit charge associated with those position. A point of higher potential would mean that a $$1$$ C charge at that position has higher energy than it would have in the power position. So,we see that in defining potential, a reference charged body is needed.

Now, in a battery there is a positive terminal and a negative terminal. Positive charge is stored in positive terminal and negative charge is stored in negative terminal. Now,a the voltage of battery being $$V$$ would mean that the positive charges have $$V$$ energy per coulomb more than the negative charges. But how are we being able to define energy and potential without a reference point in this case? The difference between potentials would exist if we calculated everything with respect to a certain electrical charge. So,with respect to which charge are we calculating the difference in electrical energy and potential of the positive and negative charges in this case?I am finding it hard to understand.

• For the battery the reference point (point considered to be at zero potential) is typically the negative terminal. Sep 8 at 15:58
• potential difference between two points is independant of the reference choice for the zero of potential. Sep 8 at 16:12
• Your statement "we need to take a charged body as reference" is not true. you could take any point in space, for theorie we usually take an uncharged point at infinity, for practical use, we take a good connection to earth. Sep 8 at 17:25
• @BobD could you please explain why that's the case? Why is the positive charge considered to be at a higher potential?Why can't we assign the higher one to negative charges since the scenario is quite symmetrical where the words positive and negative don't seem to matter. Sep 10 at 15:04
• @a_i_r I will post an answer to explain. Sep 10 at 15:19

When we define electric potential of a charge, we need to take a charged body as a reference and then we bring the test charge to calculate the electric potential.

The reference point (point assigned a potential of zero) need not be a charged body. It can be any point arbitrarily chosen (though generally logically chosen based on the analysis involved). The important thing is there needs to be an electric field between the two points in order for there to be a potential and potential difference. This follows from the definition of electric potential:

The electric potential is the amount of work required to move a unit charge from a reference point (point assigned an electric potential of zero) to a specific point against the direction of an electric field between the two points.

That the potential is based on the work done against the electric field is important because, by convention, the direction of the electric field is taken as the direction of the force that a positive charge would experience if placed in the field. That, in turn, is based on the convention in electrical engineering that current is the flow of positive charge (so called conventional current). This convention was established well before electrical current was understood to primarily consist of the flow of electrons.

Now, the voltage of battery being $$V$$ would mean that the positive charges have $$V$$ energy per coulomb more than the negative charges.

It means that it takes $$V$$ Joules of positive work to move one coulomb of positive charge from the negative battery terminal to the positive battery terminal. It is positive work because the positive charge has to be moved against the (repulsive) direction of the electric field. Positive work transfers energy to the charge increasing its potential and potential energy, relative to the negative charge.

But how are we being able to define energy and potential without a reference point in this case?

The reference point is the negative terminal of the battery. Then the potential is zero at the negative terminal and $$V$$ at the positive terminal making the potential difference $$V$$. Note that if we assigned the negative terminal a potential of 1 volt, the potential at the positive terminal would be $$V+1$$, and the potential difference would still be $$V$$.

The difference between potentials would exist if we calculated everything with respect to a certain electrical charge.

If by "certain electrical charge" you mean some other charge than at the battery terminals, then yes. The potentials at the battery terminals would be different but the potential difference would be the same.

So, with respect to which charge are we calculating the difference in electrical energy and potential of the positive and negative charges in this case?

The negative terminal of the battery, as described above.

With regard to your following comment as to why the negative terminal is considered the reference point:

could you please explain why that's the case? Why is the positive charge considered to be at a higher potential?Why can't we assign the higher one to negative charges since the scenario is quite symmetrical where the words positive and negative don't seem to matter

You're correct about the symmetry. The negative terminal would be at higher potential than the positive terminal if, by convention, the direction of the electric field was chosen as the direction of the force that a negative charge would experience if placed in the field, and current was chosen to be the flow of negative (electron) charge. The convention for current is unfortunate, but we've learned to live with it.

Hope this helps.

But how are we being able to define energy and potential without a reference point in this case?

Because any appropriate reference point disappears with subtraction.

If I calculate the energy released by a book falling 1m, it doesn't matter if I use the surface of the table, the floor of the room, or the center of the earth as the height reference. The total energy of the book is different in each case, but the energy released after dropping a meter is the same.

$$\Delta E = (100\text{g})(g)(1\text{m}) - (100\text{g})(g)(0\text{m}) = 0.98\text{J}$$ $$\Delta E = (100\text{g})(g)(2.4\text{m}) - (100\text{g})(g)(1.4\text{m}) = 0.98\text{J}$$ $$\Delta E = (100\text{g})(g)(6372\text{m}) - (100\text{g})(g)(6373\text{m}) = 0.98\text{J}$$

Likewise for the battery, I can use an arbitrary reference point for potentials, but the difference in potentials of the battery terminals is constant since the reference point is subtracted from both sides.

In the case of the book, I knew the potential energy difference per mass between the points in the gravitational field. But I had to have a specific mass to calculate a particular energy difference.

For the battery, we know the potential energy difference per charge between the battery terminals. We would have to assume a particular amount of charge to determine the amount of energy released or consumed by the motion.

Since it's more convenient to leave the amount of charge unknown, we just keep track of the potential difference per charge.