I Motion of a charged particle -- Changes in KE and PE?

[aaronll]

If a charged particle moves through a potential difference, it gains kinetic energy but does it also lose potential energy?
When I accelerate a particle and then I "free it", what happen to its potential energy if the total energy should be conserved?

 

[anuttarasammyak]

Let us see a pendulum. It has maximum kinetic energy at the bottom where potential energy is minimum. It has minimum zero kinetic energy at the top turning points where potential energy is maximum. This example suggests your thought.

 

[PeroK]

If a charged particle moves through a potential difference, it gains kinetic energy but does it also lose potential energy?

Yes, it must. It can only gain KE if it loses PE. That's the nature of a conservative force.

 

[aaronll]

Yes, it must. It can only gain KE if it loses PE. That's the nature of a conservative force.

Yes I know, maybe my question is very wrong, but if I have this apparatus:

Before: particle have a kinetic energy with v1 and so total energy = kinetic energy.
Enter: particle is due to the force, that is conservative, so E is conserved and gain kinetic energy ( loss potential energy).
flow Out: It has only kinetic energy, but is different from when is inside, because there are no more "potential" energy they have relatively to the position in the field.

I think I'm wrong because potential energy have a meaning only when the particle is in a field so there is potential difference ( this is what interest) and energy is conserved because inside the field do a work and convert that work in kinetic energy, is correct?

thanks

 

[PeroK]

If we take that diagram at face value: there are six postive charges (held in place by an external force) and likewise six negative charges. Then, we have a charged particle moving through the electric field created by those charges. That electric field does not exist only within the plates. It may be an approximation that the force only acts on the particle while it is between the plates, but if you look at the set-up fully, then there must be a potential outside the plates. In particular, a negatively charged particle must have more potential energy before it enters the plates than afterwards, if it moves as shown on your diagram.

 

[aaronll]

If we take that diagram at face value: there are six postive charges (held in place by an external force) and likewise six negative charges. Then, we have a charged particle moving through the electric field created by those charges. That electric field does not exist only within the plates. It may be an approximation that the force only acts on the particle while it is between the plates, but if you look at the set-up fully, then there must be a potential outside the plates. In particular, a negatively charged particle must have more potential energy before it enters the plates than afterwards, if it moves as shown on your diagram.

Maybe my question was not clear, I try to reformulate with a different situation:
I have a table, 6m tall relatively to the "ground", with a cube on it.
The cube has total energy of 0 ( is at rest and potential energy relatively to the table surface is 0).
So E = 0.

Now, I move the cube beyond the edge of the table, and now it is 6m from the ground ( also before but that is not my doubt) so its energy is the potential energy relatively to the ground, so E > 0.

The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.
How It is?

 

[vanhees71]

I think it's simpler to get the equations right first. Having a "conservative force" means that the force can be written as the gradient of a scalar potential, and everything is not explicitly time-dependent
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Your capacitor at a static voltage is an example. Another one is the gravitational force between some very light object and other very heavy objects at rest.

From Newton's equation of motion,
$$m \ddot{\vec{x}}=\vec{F}(\vec{x}) = -\vec{\nabla} V(\vec{x}),$$
you can then derive the energy-conservation law by multiplying with $\dot{\vec{x}}$ and integrating between two arbitrary times. On the left-hand side this gives
$$\int_{t_1}^{t_2} \mathrm{d} t m \dot{\vec{x}} \cdot \ddot{\vec{x}} = \int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right) = \left [\frac{m}{2} \dot{\vec{x}}(t) \right ]_{t=t_1}^{t=t_2} = T(t_2)-T(t_1),$$
where the kinetic energy at time $t$ is defined as
$$T(t)=\frac{m}{2} \dot{\vec{x}}^2(t).$$
On the right-hand side the above procedure gives
$$\int_{t_1}^{t_2} \mathrm{d} t \mathrm{d} t [-\dot{\vec{x}} \cdot \vec{\nabla} V] = -\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} V = - \left \{ V[\vec{x}(t_2)] - V[\vec{x}(t_1)] \right \}.$$
Setting this equal to the previous equation and sorting the quantities at each time you get
$$T(t_2)+V[\vec{x}(t_2)]=T(t_1)+V[\vec{x}(t_1)] \; \rightarrow \; E=T+V=\text{const}$$
along the trajectory.

That energy-conservation law is very useful, because you don't need to know the solution of the equations of motion to calculate the energy, provided the force has a potential. For a given force a potential exists if the line integral of the force between two points is independent on the path connecting these two points, and then the potential is given by the line integrals connecting some fixed point with any other point,
$$V(\vec{x})=-\int_{C_{\vec{x}_0 \rightarrow \vec{x}}} \mathrm{d} \vec{x}' \cdot \vec{F}(\vec{x}').$$

Now if you have the situation that $V(\vec{x}) \rightarrow 0$ for $r=|\vec{x}| \rightarrow \infty$ and the particle enters with some velocity $\vec{v}$ from infinity, it gets scattered by the force described by the potential and then goes to infinity again. Then far away from the region, where the potential is non-zero it has some speed $\vec{v}'$, but since $V=0$ in the beginning and at the end of the scattering process the energy-conservation law tells you that the speed is the same initially and finally $|\vec{v}'|=|\vec{v}|$.

 

[anuttarasammyak]

The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.
How It is?

In comparison of before and after, energy ground level should be in common. If you can change energy ground level any time you like, I cannot say anything right for you without knowing your rule.

 

[PeroK]

The value of energy is changed ( value in joules ) but only because I change the "reference", so energy seems to be not conserved, or change value when I change the position of the object but doing no work.
How It is?

Conserved means a quantity that does not change over time. Energy of a closed system is conserved. Energy, however, is frame dependent. There is no absolute value for the energy of a system. If you change your reference point for PE, then the energy of the system changes from one reference frame to another.

However, the change in energy of a system should be the same in any reference frame. Wherever you choose your reference point for zero PE, the change in PE when a particle moves over time is the same.

 

[jartsa]

Yes I know, maybe my question is very wrong, but if I have this apparatus:
View attachment 291013
Before: particle have a kinetic energy with v1 and so total energy = kinetic energy.
Enter: particle is due to the force, that is conservative, so E is conserved and gain kinetic energy ( loss potential energy).
flow Out: It has only kinetic energy, but is different from when is inside, because there are no more "potential" energy they have relatively to the position in the field.

I think I'm wrong because potential energy have a meaning only when the particle is in a field so there is potential difference ( this is what interest) and energy is conserved because inside the field do a work and convert that work in kinetic energy, is correct?

thanks

You did not draw the important part. I mean the line ends where the particle is not yet free.