# Energy due to a charge is infinite?

• A Dhingra
In summary: This is incorrect. Simply because the distance is infinite does not imply that the energy is infinite. Work is W=∫F.dx . So in it if i apply the limits some distance r to ∞, i would get,W = ∫(kQq/x2) dx = -kQq/x, x from r to ∞, = -kqQ(1/∞-1/r) = -kQq/r.This is definitely not ∞.
A Dhingra
energy due to a charge is infinite??

hello..

My professor said "Law of conservation of energy is not applicable to waves." This puzzled me a lot and trying to think otherwise i have a doubt regarding energy due to electric field. It seems that if energy due to electric field and magnetic combined, this might turn out to be true for light.
Let us have a charge Q at rest, whose electric field exists in the whole space. If a test charge,q i.e. Q>>q, is placed in this region, then due to repulsion it will move away from the charge Q. It will continue this motion till it is at infinite distance. This suggests that the total energy gained by the charge q from Q is infinite, further suggesting that the energy due to the electric field is not getting used up... i.e. i can bring as many test charges as i want but this repulsion will exist, no matter how much energy the previous one has taken. So how is that so?

What kind of energy is possessed by the electric field created by a charge? Is it never ending, and suggesting that it never got created (may be when the charge or the electric field was formed)? in that sense this need not obey the laws of conservation of energy!

may be what i have asked is very silly, but any help is appreciated..

A Dhingra said:
If a test charge,q i.e. Q>>q, is placed in this region, then due to repulsion it will move away from the charge Q. It will continue this motion till it is at infinite distance. This suggests that the total energy gained by the charge q from Q is infinite
This is incorrect. Simply because the distance is infinite does not imply that the energy is infinite.

What is the general expression for work? What does the energy depend on besides the distance? What happens if this other quantity goes to 0 and is that the situation here?

Welcome to PF;

What you want to investigate is how much energy it takes to accelerate a charged particle compared with a neutral particle.
[Dalespam has shown you the way to challenge the argument you were provided with.]

Work is W=∫F.dx .
So in it if i apply the limits some distance r to ∞, i would get,
W = ∫(kQq/x2) dx = -kQq/x, x from r to ∞,
= -kqQ(1/∞-1/r) = -kQq/r..
this is definitely not ∞.

But I was thinking that the test charge is moving continuously (with non uniform, decreasing acceleration) almost for infinite time. In that case the sum of all the small patches under the curve should have given infinite energy(I was considering kinetic energy, on v-t graph)...but that doesn't seem to be happening...

A Dhingra said:
hello..

Let us have a charge Q at rest, whose electric field exists in the whole space. If a test charge,q i.e. Q>>q, is placed in this region, then due to repulsion it will move away from the charge Q. It will continue this motion till it is at infinite distance. This suggests that the total energy gained by the charge q from Q is infinite, further suggesting that the energy due to the electric field is not getting used up...

As was already pointed out to you(by Dalespam) this process does not render an infinite amount of energy but besides that this energy is not produced by the rest charge Q alone, this energy is assigned to the system of two charges and it is yust the energy recovered from the one used to assemble the two charges.
However there is a problem if you think of the energy of the charge Q alone since this energy is indeed infinite. This is a known problem of electrodynamics and is an indication that classical electrodynamics is not logically consistent unlike Newtonian mechanics which does not suffer from consistency problems.

A Dhingra said:
Work is W=∫F.dx .
So in it if i apply the limits some distance r to ∞, i would get,
W = ∫(kQq/x2) dx = -kQq/x, x from r to ∞,
= -kqQ(1/∞-1/r) = -kQq/r..
this is definitely not ∞.
Exactly! Well done.

A Dhingra said:
But I was thinking that the test charge is moving continuously (with non uniform, decreasing acceleration) almost for infinite time. In that case the sum of all the small patches under the curve should have given infinite energy(I was considering kinetic energy, on v-t graph)...but that doesn't seem to be happening...
Again, you need to carry out the appropriate integral. However, you seem to think that the area under a v-t graph represents energy. It does not.

Think about what the appropriate integrand is, consider impulse, momentum, and the relationship between momentum and KE. Look at that integrand, is there some part of it that makes you think it might not be infinite? If you can, perform the integral.

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You must have misquoted your professor or taken eir words out of context. Law of conservation of energy applies everywhere, except possibly on a cosmological scale. (The latter uncertainty is due to our lack of understanding of gravity, ad hoc inflaton fields, and dark energy.)

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Khashishi said:
You must have misquoted your professor or taken eir words out of context. Law of conservation of energy applies everywhere...
Or the prof is using a pedagogical technique hoping to stimulate inquiry, a bit like the famous goldfish problem (why does a goldfish not displace any water?) ... the statement itself looks like an oblique reference to what facenian was talking about.
However there is a problem if you think of the energy of the charge Q alone since this energy is indeed infinite. This is a known problem of electrodynamics and is an indication that classical electrodynamics is not logically consistent

## 1. What is meant by "energy due to a charge is infinite"?

When a charged particle is placed in an electric field, it experiences a force that causes it to move. This movement requires energy, and the amount of energy required is directly proportional to the strength of the electric field. In some cases, the electric field can have an infinite strength, leading to an infinite amount of energy needed to move the charged particle.

## 2. Is it possible for an electric field to have an infinite strength?

In theory, yes. An infinite electric field can be produced by an infinitely large and infinitely charged object. However, in reality, such extreme conditions are not possible as they violate the laws of physics.

## 3. Does this mean that the energy of a charged particle in an electric field is also infinite?

No, the energy of a charged particle is not infinite. While the electric field may have an infinite strength, the energy of the charged particle is limited by its mass. As the particle approaches the speed of light, its energy also approaches a maximum value known as the rest energy.

## 4. How does this concept of infinite energy due to a charge relate to real-world applications?

In most real-world scenarios, the electric field strength is not infinite, so the energy of a charged particle is finite. However, this concept is crucial in understanding the behavior of particles in high-energy environments, such as in particle accelerators, where the electric field strength can approach extremely high levels.

## 5. Can the concept of infinite energy due to a charge be applied to other types of fields?

Yes, similar concepts can be applied to other types of fields, such as magnetic fields. In some cases, the magnetic field strength can also reach infinite levels, leading to infinite energy requirements for moving charged particles. This concept is important in understanding the behavior of particles in magnetic confinement systems, such as in fusion reactors.

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