# B How can we talk about charges very close to zero?

1. Oct 3, 2016

### Prem1998

I was reading about the definition of electric field. It said that electric field at a point due to a source charge is the force exerted by it on another charge placed at that point divided by the magnitude of charge placed at that point.
But, obviously, if we place any finite charge in electric filed, it will disturb the original source charge and thereby it will change the value of electric field at that point.
So, when defining electric field we often use the term 'test charge' which is infinitesimally small so that it does not disturb the original charge. Mathematically we represent it as lim F/q as q tends to zero.
But we cannot talk about such a charge given that charge is quantized. The smallest building block of charge which I'm aware of is a quark. And, it is still something finite so that it will disturb the source charge. Then, how can we talk about charges which are as close to zero as possible? Doesn't this seem like implying that charge distributions are continuous and charge can take any value like mass?

2. Oct 4, 2016

### Simon Bridge

We use a model that allows that.
Yes.

The "test charge" is entirely imaginary.
Mathematics does not have to strictly follow reality to be useful - we can use the continuous definition to work with quantized charges.

3. Oct 4, 2016

### Prem1998

Thanks for the reply. So, it's just a mathematical way to find the value of electric field and does't imply that charges are continuous.
But could it be that charge can also be arbitrary like masses? I mean, first we used to think of electrons as the quantum, now we have quarks. The size is getting smaller. Also, is charge still something really fundamental or has the cause of charge been discovered? Like what it is and it's relation with mass.

4. Oct 4, 2016

### houlahound

As our knowledge increases our technology gets better which increases our knowledge to make better technology.....so yes we can probe deeper into the very small and extremely vast and uncover more details.

Will this process ever reach a limit, nobody knows.

5. Oct 4, 2016

### PeroK

To expand on what Simon says. There are two ways to interpret an idealised mathematical model:

a) To take the assumption literally. In this case that the test charge does not affect the electric field.
b) That the assumption has no significant effect on the result. In this case, that the effect of the test charge on the elecric field is negligible (quantitatively).

It's important to know when to use one model and when to use another. A good example is the assumption that the gravitational field near the surface of the Earth remains constant with height. Clearly it doesn't, but for many calculations the difference between a constant field and a field that changes with height is negligible. For example, the difference in time taken for an object to fall from 1000m using these two models is about 0.002s. If this time is significant, then you would have to use the more complicated varying gravity model. But for most practical purposes the simple constant gravity model will do.

Note that in both cases you can assume the Earth does not move - or, at least, moves to an utterly negligible degree.

If, however, you are studying planetary orbits, the first assumption (gravity constant with height) must definitely be dropped and possibly the second assumption (that the more massive object does not move) may have to be dropped as well, depending on the accuracy you require of your calculations.

6. Oct 4, 2016

### Prem1998

I think that in this case the mathematical model also wants us to take the assumptions literally. I mean, that's why, the actual formula for calculation also says that q tends to zero. Note that the charge on a quark is a number very far from zero from the mathematical point of view. Whatever number you take to be very close to zero, it's still very far and still a smaller one is possible. It's just that the difference is not significant for us. But, for a very very small mass, the force exerted by a quark could cause significant acceleration to it and thereby could disturb it's position significantly. It is for the same reason for which we can't use Earth as the test mass to calculate Gravitational field due to the sun but we can take our bodies to be a test mass to calculate the gravitational field due to Earth. I agree with you in the case of acceleration due to gravity near the surface of the Earth.
But, in this case, maybe it's better that we take the formula literally. After all, it's an assumption just for a calculation to determine the electric field. After we get the electric field, the assumption has nothing to do with reality.
And, maybe, charge is also continuous. The quantum size is getting smaller with further research.

7. Oct 4, 2016

### PeroK

Actually, that's exactly how you would calculate the Earth's orbit round the Sun: assume that the effect of the Earth's gravity is negligible compared to the Sun's and treat the Earth as a test mass. This works in both classical and GR formulations of gravity.

8. Oct 4, 2016

### Prem1998

What? Is it really true? I'm sorry if I made wrong statements. But, is the mass of the sun really that large to neglect the high gravitational force between Earth and Sun?

9. Oct 4, 2016

### PeroK

I'm afraid so! I can quote from Hartle's "Gravity": "These test particles might be the planets orbiting our Sun ..."

10. Oct 4, 2016

### Prem1998

Then, I would like to revise my statement to this:
'We can't assume another star to be a test mass to calculate gravitational field due to the sun'. I think whether changes are significant or not is just about whether the masses and charges are comparable or not.
And, I have this one more reason to support that the formula has to be interpreted literally:
Usually, we are told that electric field is force exerted per unit charge. But this is a formula more than a definition. And, in case of this definition, it's true that we can't interpret the ideal situation literally. It's because whatever charge is placed in the field, it always disturbs the source charge. So, in this case, we have to talk about whether the changes are significant or not.
BUT I like to think of electric filed more like this: And electric field is just a modification of space with numbers defined at every point. And, in order to get to those numbers, we have to find the answer to the hypothetical question that 'what force per unit charge will be experienced by a charge if it is placed at that point?', the formula helps us to get to that number by assuming what would be the value of electric field if no charge was present at the point to disturb the source charge?

11. Oct 4, 2016

### Prem1998

Actually, you're right. I now understand that this argument is really worthless. It doesn't change things whether we bring a quark in an electric field or we bring a hypothetical test charge in an electric field. I'm sorry for wasting your time.

12. Oct 4, 2016

### Simon Bridge

It's not a waste of time when it means someone comes to a new understanding ... if you peruse the threads you'll see we sometimes just tell people to stop being stupid - though not in so many words. When we spend some effort on someone's question it is because that person shows they are prepared to think about things in a way that will also be helpful to other people reading the discussion years later.
You are doing good: keep thinking.