0
$\begingroup$

How to understand the concept of work using the following example:

a positively charged particle (with charge $Q$) is placed in between a charged in a homogeneous electric field between the capacitor plates. This charged particle is placed close to the positive plate.

dotted charge denotes the end position of the charge at P2

When this particle is released at P1, then the electric force acting upon it (to the right) will cause it to accelerate to the right. Therefore its kinetic energy will increase. This tells me that the charge had a greater Potential Energy at the start (P1).

-

However I am confronted with the idea that the system as a whole has done negative work, because we did not do anything (no exterior influences/forces) in order to move the particle. Could someone explain to me why this is the case?

$\endgroup$
0
$\begingroup$

It is not unlike dropping a mass in a constant gravity field.

The force applied is constant (ideally) and the work is $F.dx$ in module (in the electrical example as well as in the gravitational example).

In both cases, the object has lost potenttial energy and gained kinetic energy and the overall balance remains $0$.

$\endgroup$
2
  • $\begingroup$ May I ask, if a moving charge decreases or changes in anyway the energy stored in the capacitor. (Assuming the capcitor is charged and disconnected from the battery) $\endgroup$ – Arman Braca Jun 9 at 14:20
  • $\begingroup$ Only if said charge which has been set in motion does some work, for instance if it collides with a gas between the electrodes, otherwise said charge in motion is the equivalent of a (tiny) current $i$ meeting no resistance whose work is then symbolically $R.i^{2}$ ( which is $0$ if R is $0$). $\endgroup$ – Serge Hulne Jun 9 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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