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Energy is defined as the capacity to do work. Work in turn is defined as force x displacement.

An electric field exerts the field in all directions infinitely (even though the strength of that force will be lower as we move away from the source).

Since the field can extend till infinity, it can exert a force on the charge and move it to an infinite distance (theoretically).

Is the energy on an electric field infinite by this logic?

If so, wouldn't that mean the energy in a battery is infinite?

Is the definition of energy in terms of force x displacement related to the energy in terms of heat, electricity, chemical bonding and other sources?

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  • $\begingroup$ The language of physics is mathematics. not words. Words are necessary for definitions but are useless for calculating solid numbers as you are doing getting the idea that the energy in the field is infinite. If one puts down the numbers in the formulas, this is wrong. $\endgroup$
    – anna v
    Commented Apr 30, 2022 at 9:47
  • $\begingroup$ The amount of work done would be a non terminating number. It would not be infinite as you think. As you go far from source of charge, the work done gets smaller and smaller per unit length. It is similar to a number whose value is increasing but decimal expansion is also increasing continuously with that. But the value of that number is in a range of finite numbers. You should have checked the calculations once. $\endgroup$
    – James Webb
    Commented Apr 30, 2022 at 12:20

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"Can move it an infinite distance" is no way to measure energy. Give an object any amount of kinetic energy, and in the presence of no other forces, It will move an infinite distance.

$\int_{r}^{\infty} \vec{E} \cdot \vec{dl}$

Represents the amount of work done by the electric field in moving an object from r to $\infty$

For a point charge, For all r ≠ 0, this qauntity is finite.

Probably not what your asking :

The formula for electric field energy

$\iiint \frac{1}{2}\epsilon_0 |\vec{E}|^2 dv $

Is derived by finding the amount of work to assemble a charge distribution

This is derived from the discrete version of this formula, finds the work to build up a distribution of point charges. Deriving this formula needs you to disregard the potential of the charge you're building up, at that particular moment in time, ( since it does not repell itself)

In deriving the continuous version, the generalisation to a distribution ignores this condition since each element $\rho dv$'s potential is zero for a finite $\rho$. ( as discussed in griffiths) so it yields the same result either way.

However, For a point charge $\rho = Q\delta^3(r)$

This is infinite, so our formula somewhat breaks down for point charges

This leads to an incorrect reading that the energy of a single point charge is infinity.

This result of infinities when dealing with problematic point charges is often manually subtracted, as the formula is only meant to find the potential energy of the distribution, and not the added infinities.

Modeling charges a spherical balls of charge with a finite radius however fixes this problem and allows the formula to work without ignoring certain assumptions

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  • $\begingroup$ Agreed that moving an infinite distance alone does not mean energy is added to it. But if a force is acting on an object in the same direction of motion continuously while it is moving, then by definition, work is being done on it by that force. $\endgroup$
    – Adarsh
    Commented Apr 30, 2022 at 10:39
  • $\begingroup$ Correct, but just because work is being done on it over an infinite distance, doesn't nesseccerily imply the work done is divergent. The sequence 1/r^2 converges $\endgroup$
    – jensen paull
    Commented Apr 30, 2022 at 12:38
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Is the energy on an electric field infinite by this logic?

No. Not all functions that extend to infinity have an infinite integral. In particular for $$W = \int_{r_0}^{\infty} f(r) \ dr$$ $W$ is only infinite if $f(r)$ goes to zero slower than $1/r$.

By Coulomb’s law $$f(r)=k\frac{q_1 q_2}{r^2}$$ goes to zero faster than $1/r$ so the work is finite when going from some finite distance to infinity.

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