1
$\begingroup$

My textbook reads that when a mass/ charge is moved into the influence of a gravitational/ electric field the field gains/ loses energy rather than the mass/ charge. How can this be explained?

$\endgroup$

1 Answer 1

1
$\begingroup$

My textbook reads that when a mass/ charge is moved into the influence of a gravitational/ electric field the field gains/ loses energy rather than the mass/ charge. How can this be explained?

$\let\eps=\varepsilon \let\sig=\sigma \def\half{{\textstyle{1 \over 2}}} \def\cE{\mathcal E}$ In a sense they are both true. They are alternative descriptions of the same phenomenon. Sometimes one is more useful, sometimes the other.

Consider e.g. the gravitational case, in the framework of Newtonian physics. Although it would be possible to think of energy of the gravitational field, when it comes to computing planets' motions nobody does it. Instead it helps very much to attribute a gravitational energy to Sun and planets themselves, as a function of their reciprocal distances.

In e.m. case both viewpoints may be of use, according to the problem. Consider e.g. a plane capacitor. Then it's easier to think of energy as due to the electric field between plates. Maybe you know the formula: $$\cE = \half \eps_0 E^2 V \tag1$$ where $V = S\,d$ is volume of space between plates (I assumed a uniform field).

We may play with eq. (1), using the espression for electric field $$E = {\sig \over \eps_0}.$$ We find $$\cE = \half\,\eps_0\,S\,d\,{\sig^2 \over \eps_0^2} = {Q^2 d \over 2\,\eps_0\,S} = {Q^2 \over 2\,C}.\tag2$$ I've used $Q = \sig\,S$ and capacity formula: $$C = {\eps_0\,S \over d}.\tag3$$

Remember that eq. (2) is usually arrived at by another way, so this is a useful check. But one could turn the argument around, deriving ffirst (3), then energy (2), and arriving at eq. (1) for energy as a function of field.

Viewing energy as a property of fields is also important in more complex cases but I don't write on that because I'm not sure you heard of the relevant physics (e.g. e.m. waves).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.